mysql_fetch_array（）/ mysql_fetch_assoc（）/ mysql_fetch_row（）/ mysql_num_rowsなど…パラメータ1がリソースであることが必要です

Question

MySQLテーブルからデータを選択しようとしていますが、次のいずれかのエラーメッセージが表示されます。

mysql_fetch_array（）は、パラメーター1がリソースであり、ブール値が指定されていることを想定しています。

これは私のコードです：

$username = $_POST['username'];
$password = $_POST['password'];

$result = mysql_query('SELECT * FROM Users WHERE UserName LIKE $username');

while($row = mysql_fetch_array($result)) {
    echo $row['FirstName'];
}

Answer 1

クエリはさまざまな理由で失敗する可能性があります。その場合、mysql_ *とmysqli拡張機能の両方falseがそれぞれのクエリ関数/メソッドから返されます。そのエラー状態をテストし、それに応じて処理する必要があります。

mysql_ *拡張子：

NOTE mysql_関数は廃止されましたし、PHPのバージョン7で削除されました。

$resultに渡す前に確認してくださいmysql_fetch_array。falseクエリが失敗したことが原因であることがわかります。mysql_query考えられる戻り値とそれらの処理方法に関する提案については、ドキュメントを参照してください。

$username = mysql_real_escape_string($_POST['username']); $password = $_POST['password']; $result = mysql_query("SELECT * FROM Users WHERE UserName LIKE '$username'"); if($result === FALSE) { 
    die(mysql_error()); // TODO: better error handling
}

while($row = mysql_fetch_array($result))
{
    echo $row['FirstName'];
}

mysqli拡張の
手続き型：

$username = mysqli_real_escape_string($mysqli, $_POST['username']);
$result = mysqli_query($mysqli, "SELECT * FROM Users WHERE UserName LIKE '$username'"); // mysqli_query returns false if something went wrong with the query if($result === FALSE) { 
    yourErrorHandler(mysqli_error($mysqli)); } else { // as of php 5.4 mysqli_result implements Traversable, so you can use it with foreach foreach( $result as $row ) {
        ...

ooスタイル：

$username = $mysqli->escape_string($_POST['username']);
$result = $mysqli->query("SELECT * FROM Users WHERE UserName LIKE '$username'"); if($result === FALSE) { 
    yourErrorHandler($mysqli->error); // or $mysqli->error_list
}
else {
    // as of php 5.4 mysqli_result implements Traversable, so you can use it with foreach
    foreach( $result as $row ) {
      ...

プリペアドステートメントの使用：

$stmt = $mysqli->prepare('SELECT * FROM Users WHERE UserName LIKE ?');
if ( !$stmt ) { yourErrorHandler($mysqli->error); // or $mysqli->error_list } else if ( !$stmt->bind_param('s', $_POST['username']) ) { yourErrorHandler($stmt->error); // or $stmt->error_list } else if ( !$stmt->execute() ) {
    yourErrorHandler($stmt->error); // or $stmt->error_list
}
else {
    $result = $stmt->get_result();
    // as of php 5.4 mysqli_result implements Traversable, so you can use it with foreach
    foreach( $result as $row ) {
      ...

---

これらの例は、実行する必要があること（エラー処理）を示しているだけであり、その方法は示していません。or dieHTMLを出力するときにプロダクションコードを使用しないでください。使用しないと、（少なくとも）無効なHTMLが生成されます。また、データベースのエラーメッセージは、情報が多すぎるため、管理者以外のユーザーには表示しないでください。

Answer 2

This error message is displayed when you have an error in your query which caused it to fail. It will manifest itself when using:

• mysql_fetch_array/mysqli_fetch_array()
• mysql_fetch_assoc()/mysqli_fetch_assoc()
• mysql_num_rows()/mysqli_num_rows()

Note: This error does not appear if no rows are affected by your query. Only a query with an invalid syntax will generate this error.

Troubleshooting Steps

• Make sure you have your development server configured to display all errors. You can do this by placing this at the top of your files or in your config file: error_reporting(-1);. If you have any syntax errors this will point them out to you.

• Use mysql_error(). mysql_error() will report any errors MySQL encountered while performing your query.

  Sample usage:

  mysql_connect($host, $username, $password) or die("cannot connect"); mysql_select_db($db_name) or die("cannot select DB");
  
  $sql = "SELECT * FROM table_name"; $result = mysql_query($sql); if (false === $result) {
      echo mysql_error();
  }
  

• Run your query from the MySQL command line or a tool like phpMyAdmin. If you have a syntax error in your query this will tell you what it is.

• Make sure your quotes are correct. A missing quote around the query or a value can cause a query to fail.

• Make sure you are escaping your values. Quotes in your query can cause a query to fail (and also leave you open to SQL injections). Use mysql_real_escape_string() to escape your input.

• Make sure you are not mixing mysqli_* and mysql_* functions. They are not the same thing and cannot be used together. (If you're going to choose one or the other stick with mysqli_*. See below for why.)

Other tips

mysql_* functions should not be used for new code. They are no longer maintained and the community has begun the deprecation process. Instead you should learn about prepared statements and use either PDO or MySQLi. If you can't decide, this article will help to choose. If you care to learn, here is good PDO tutorial.

Answer 3

Error occurred here was due to the use of single quotes ('). You can put your query like this:

mysql_query("
SELECT * FROM Users 
WHERE UserName 
LIKE '".mysql_real_escape_string ($username)."'
");

It's using mysql_real_escape_string for prevention of SQL injection. Though we should use MySQLi or PDO_MYSQL extension for upgraded version of PHP (PHP 5.5.0 and later), but for older versions mysql_real_escape_string will do the trick.

Answer 4

As scompt.com explained, the query might fail. Use this code the get the error of the query or the correct result:

$username = $_POST['username'];
$password = $_POST['password'];

$result = mysql_query(" SELECT * FROM Users WHERE UserName LIKE '".mysql_real_escape_string($username)."'
");

if($result) { while($row = mysql_fetch_array($result)) { echo $row['FirstName'];
    }
} else {
    echo 'Invalid query: ' . mysql_error() . "\n";
    echo 'Whole query: ' . $query; 
}

See the documentation for mysql_query() for further information.

The actual error was the single quotes so that the variable $username was not parsed. But you should really use mysql_real_escape_string($username) to avoid SQL injections.

Answer 5

Put quotes around $username. String values, as opposed to numeric values, must be enclosed in quotes.

$result = mysql_query("SELECT * FROM Users WHERE UserName LIKE '$username'");

Also, there is no point in using the LIKE condition if you're not using wildcards: if you need an exact match use = instead of LIKE.

Answer 6

Please check once the database selected are not because some times database is not selected

Check

mysql_select_db('database name ')or DIE('Database name is not available!');

before MySQL query and then go to next step

$result = mysql_query('SELECT * FROM Users WHERE UserName LIKE $username');

f($result === FALSE) {
    die(mysql_error());

Answer 7

Your code should be something like this

$username = $_POST['username'];
$password = $_POST['password'];
$query = "SELECT * FROM Users WHERE UserName LIKE '$username'";
echo $query; $result = mysql_query($query); if($result === FALSE) {
    die(mysql_error("error message for the user")); 
}

while($row = mysql_fetch_array($result))
{
    echo $row['FirstName'];
}

Once done with that, you would get the query printed on the screen. Try this query on your server and see if it produces the desired results. Most of the times the error is in the query. Rest of the code is correct.

Answer 8

$result = mysql_query('SELECT * FROM Users WHERE UserName LIKE $username');

You define the string using single quotes and PHP does not parse single quote delimited strings. In order to obtain variable interpolation you will need to use double quotes OR string concatenation (or a combination there of). See http://php.net/manual/en/language.types.string.php for more information.

Also you should check that mysql_query returned a valid result resource, otherwise fetch_*, num_rows, etc will not work on the result as is not a result! IE:

$username = $_POST['username'];
$password = $_POST['password'];
$result = mysql_query('SELECT * FROM Users WHERE UserName LIKE $username');

if( $result === FALSE ) { trigger_error('Query failed returning error: '. mysql_error(),E_USER_ERROR); } else { while( $row = mysql_fetch_array($result) ) { echo $row['username'];
   }
}

http://us.php.net/manual/en/function.mysql-query.php for more information.

Answer 9

This query should work:

$result = mysql_query("SELECT * FROM Users WHERE UserName LIKE '%$username%'");
while($row = mysql_fetch_array($result))
{
    echo $row['FirstName'];
}

The problem is single quotes, thus your query fails and returns FALSE and your WHILE loop can't execute. Using % allows you to match any results containing your string (such as SomeText-$username-SomeText).

This is simply an answer to your question, you should implement stuff mentioned in the other posts: error handling, use escape strings (users can type anything into the field, and you MUST make sure it is not arbitrary code), use PDO instead mysql_connect which is now depricated.

Answer 10

$username = $_POST['username'];
$password = $_POST['password'];
$result = mysql_query("SELECT * FROM Users WHERE UserName LIKE '%$username%'") or die(mysql_error());

while($row = mysql_fetch_array($result))
{
    echo $row['FirstName'];
}

Sometimes suppressing the query as @mysql_query(your query);

Answer 11

If you tried everything here, and it does not work, you might want to check your MySQL database collation. Mine was set to to a Swedish collation. Then I changed it to utf8_general_ci and everything just clicked into gear.

Answer 12

$query = "SELECT Name,Mobile,Website,Rating FROM grand_table order by 4"; while( $data = mysql_fetch_array($query)) { echo("<tr><td>$data[0]</td><td>$data[1]</td><td>$data[2]</td><td>$data[3]</td></tr>");      
}

Instead of using a WHERE query, you can use this ORDER BY query. It's far better than this for use of a query.

I have done this query and am getting no errors like parameter or boolean.

Answer 13

Try this, it must be work, otherwise you need to print the error to specify your problem

$username = $_POST['username'];
$password = $_POST['password'];

$sql = "SELECT * from Users WHERE UserName LIKE '$username'";
$result = mysql_query($sql,$con); while($row = mysql_fetch_array($result)) { echo $row['FirstName'];
}

Answer 14

There might be two reasons:

1. Have you opened the connection to the database prior to calling mysql_query function? I don't see that in your code. Use mysql_connect before making the query. See php.net/manual/en/function.mysql-connect.php

2. The variable $username is used inside a single quote string, so its value will not be evaluated inside the query. The query will definitely fail.

Thirdly, the structure of query is prone to SQL injection. You may use prepared statements to avoid this security threat.

Answer 15

Try the following code. It may work fine.

$username = $_POST['username'];
$password = $_POST['password'];
$result = mysql_query("SELECT * FROM Users WHERE UserName ='$username'");

while($row = mysql_fetch_array($result))
{
    echo $row['FirstName'];
}

Answer 16

Go to your config.php. I had the same problem. Verify the username and the password, and also sql select is the same name as the config.

Answer 17

Don't use the depricated mysql_* function (depricated in php 5.5 will be removed in php 7). and you can make this with mysqli or pdo

here is the complete select query

<?php
$servername = "localhost"; $username = "username";
$password = "password"; $dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname); // Check connection if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error); } $sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = $conn->query($sql); if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        // code here 
    }
} else {
    echo "0 results";
}
$conn->close();
?>

Answer 18

<?php
    $username = $_POST['username'];
    $password = $_POST['password'];
    $result = mysql_query("SELECT * FROM Users WHERE UserName LIKE '".$username."'");

    while($row = mysql_fetch_array($result))
    {
        echo $row['FirstName'];
    }
?>

And if there is a user with a unique user name, you can use "=" for that. There is no need to like.

Your query will be:

mysql_query("SELECT * FROM Users WHERE UserName ='".$username."'");

Answer 19

Include a connection string variable before the MySQL query. For example, $connt in this code:

$results = mysql_query($connt, "SELECT * FROM users");

Answer 20

<?php
      $username = $_POST['username'];
       $password = $_POST['password'];

     $result = mysql_query("SELECT * FROM Users WHERE UserName LIKE '".mysql_real_escape_string($username)."'")or die(mysql_error());
while($row=mysql_fetch_array($result))
  {
 echo $row['FirstName'];
 }
 ?>

Answer 21

Try This

$username = $_POST['username'];
$password = $_POST['password'];
$result = mysqli_query('SELECT * FROM Users WHERE UserName LIKE $username');

if($result){ while($row = mysqli_fetch_array($result)) { echo $row['FirstName'];
}
}

Answer 22

Any time you get the...

"Warning: mysqli_fetch_object() expects parameter 1 to be mysqli_result, boolean given"

...it is likely because there is an issue with your query. The prepare() or query() might return FALSE (a Boolean), but this generic failure message doesn't leave you much in the way of clues. How do you find out what is wrong with your query? You ask!

First of all, make sure error reporting is turned on and visible: add these two lines to the top of your file(s) right after your opening <?php tag:

error_reporting(E_ALL);
ini_set('display_errors', 1);

If your error reporting has been set in the php.ini you won't have to worry about this. Just make sure you handle errors gracefully and never reveal the true cause of any issues to your users. Revealing the true cause to the public can be a gold engraved invitation for those wanting to harm your sites and servers. If you do not want to send errors to the browser you can always monitor your web server error logs. Log locations will vary from server to server e.g., on Ubuntu the error log is typically located at /var/log/apache2/error.log. If you're examining error logs in a Linux environment you can use tail -f /path/to/log in a console window to see errors as they occur in real-time....or as you make them.

Once you're squared away on standard error reporting adding error checking on your database connection and queries will give you much more detail about the problems going on. Have a look at this example where the column name is incorrect. First, the code which returns the generic fatal error message:

$sql = "SELECT `foo` FROM `weird_words` WHERE `definition` = ?"; $query = $mysqli->prepare($sql)); // assuming $mysqli is the connection $query->bind_param('s', $definition); $query->execute();

The error is generic and not very helpful to you in solving what is going on.

With a couple of more lines of code you can get very detailed information which you can use to solve the issue immediately. Check the prepare() statement for truthiness and if it is good you can proceed on to binding and executing.

$sql = "SELECT `foo` FROM `weird_words` WHERE `definition` = ?"; if($query = $mysqli->prepare($sql)) { // assuming $mysqli is the connection $query->bind_param('s', $definition); $query->execute();
    // any additional code you need would go here.
} else {
    $error = $mysqli->errno . ' ' . $mysqli->error; // 1054 Unknown column 'foo' in 'field list'
    // handle error
}

If something is wrong you can spit out an error message which takes you directly to the issue. In this case, there is no foo column in the table, solving the problem is trivial.

If you choose, you can include this checking in a function or class and extend it by handling the errors gracefully as mentioned previously.

Answer 23

First, check your connection to the database. Is it connected successfully or not?

If it's done, then after that I have written this code, and it works well:

if (isset($_GET['q1mrks']) && isset($_GET['marks']) && isset($_GET['qt1'])) { $Q1mrks = $_GET['q1mrks']; $marks = $_GET['marks']; $qt1 = $_GET['qt1']; $qtype_qry = mysql_query("
        SELECT *
        FROM s_questiontypes
        WHERE quetype_id = '$qt1' "); $row = mysql_fetch_assoc($qtype_qry); $qcode = $row['quetype_code']; $sq_qry = "
        SELECT *
        FROM s_question
        WHERE quetype_code = '$qcode' ORDER BY RAND() LIMIT $Q1mrks
    ";
    $sq_qry = mysql_query(" SELECT * FROM s_question WHERE quetype_code = '$qcode'
        LIMIT $Q1mrks "); while ($qrow = mysql_fetch_array($sq_qry)) { $qm = $qrow['marks'] . "<br />"; $total += $qm . "<br />"; } echo $total . "/" . $marks;
}

Answer 24

Make Sure You're Not Closing Database By using db_close() Before To Running Your Query:

If you're using multiple queries in a script even you're including other pages which contains queries or database connection, then it might be possible that at any place you use db_close() that would close your database connection so make sure you're not doing this mistake in your scripts.

Answer 25

If you don't have any MySQL Error appearing while checking, make sure that you properly created your database table. This happened to me. Look for any unwanted commas or quotes.

Answer 26

Check your connection first.

Then if you want to fetch the exact value from the database then you should write:

$username = $_POST['username'];
$password = $_POST['password'];
$result = mysql_query("SELECT * FROM Users WHERE UserName =`$usernam`");

Or you want to fetch the LIKE type of value then you should write:

$result = mysql_query("SELECT * FROM Users WHERE UserName LIKE '%$username%'");

Answer 27

You can also check wether $result is failing like so, before executing the fetch array

$username = $_POST['username']; $password = $_POST['password']; $result = mysql_query('SELECT * FROM Users WHERE UserName LIKE $username'); if(!$result)
{
     echo "error executing query: "+mysql_error(); 
}else{
       while($row = mysql_fetch_array($result))
       {
         echo $row['FirstName'];
       }
}

Answer 28

Usually an error occurs when your database conectivity fails, so be sure to connect your database or to include the database file.

include_once(db_connetc.php');

OR

// Create a connection
$connection = mysql_connect("localhost", "root", "") or die(mysql_error()); //Select database mysql_select_db("db_name", $connection) or die(mysql_error());

$employee_query = "SELECT * FROM employee WHERE `id` ='".$_POST['id']."'";

$employee_data = mysql_query($employee_query);

if (mysql_num_rows($employee_data) > 0) { while ($row = mysql_fetch_array($employee_data)){ echo $row['emp_name'];
    } // end of while loop
} // end of if

• Best practice is to run the query in sqlyog and then copy it into your page code.
• Always store your query in a variable and then echo that variable. Then pass to mysql_query($query_variable);.

Answer 29

Traditionally PHP has been tolerant to bad practice and failures in code, which makes debugging quite hard. The problem in this specific case is that both mysqli and PDO by default don't tell you, when a query failed and just return FALSE. (I will not talk about the depricated mysql extention. The support for prepared statements is reason anough to switch either to PDO or mysqli.) But you can change the default behavior of PHP to always throw exceptions when a query fails.

For PDO: Use $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

error_reporting(E_ALL);

$pdo = new PDO("mysql:host=localhost;dbname=test", "test","");
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); $result = $pdo->query('select emal from users'); $data = $result->fetchAll();

This will show you the following:

Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42S22]: Column not found: 1054 Unknown column 'emal' in 'field list'' in E:\htdocs\test\mysql_errors\pdo.php on line 8

PDOException: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'emal' in 'field list' in E:\htdocs\test\mysql_errors\pdo.php on line 8

As you see, it tells you exactly, what is wrong with the query, and where to fix it in your code.

Without $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); you will get

Fatal error: Call to a member function fetchAll() on boolean in E:\htdocs\test\mysql_errors\pdo.php on line 9

For mysqli: Use mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

error_reporting(E_ALL);

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$mysqli = new mysqli('localhost', 'test', '', 'test'); $result = $mysqli->query('select emal from users'); $data = $result->fetch_all();

You will get

Fatal error: Uncaught exception 'mysqli_sql_exception' with message 'Unknown column 'emal' in 'field list'' in E:\htdocs\test\mysql_errors\mysqli.php on line 8

mysqli_sql_exception: Unknown column 'emal' in 'field list' in E:\htdocs\test\mysql_errors\mysqli.php on line 8

Without mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT); you only get

Fatal error: Call to a member function fetch_all() on boolean in E:\htdocs\test\mysql_errors\mysqli.php on line 10

Of course, you could manually check the MySQL errors. But I would go crazy if I had to do that every time I made a typo - or worse - every time I want to query the database.

Answer 30

Try this code it work fine

assign the post variable to the variable

   $username = $_POST['uname'];

   $password = $_POST['pass'];

  $result = mysql_query('SELECT * FROM userData WHERE UserName LIKE $username');

if(!empty($result)){ while($row = mysql_fetch_array($result)){ echo $row['FirstName'];
     }
}