<?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'];
    }
?>

如果有一个用户具有唯一的用户名，您可以使用“=”表示。没有必要喜欢。

您的查询将是:

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

当查询中出现错误导致查询失败时，将显示此错误消息。它会在使用时显现出来:

mysql_fetch_array / mysqli_fetch_array () 作用是()/ mysqli_fetch_assoc () mysql_num_rows () / mysqli_num_rows ()

注意:如果查询不影响任何行，则不会出现此错误。只有语法无效的查询才会产生此错误。

故障排除步骤

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.)

其他技巧

Mysql_ *函数不应该用于新代码。它们不再被维护，社区已经开始了弃用过程。相反，你应该学习准备语句并使用PDO或MySQLi。如果你不能决定，这篇文章将帮助你选择。如果你想学习，这里有一个很好的PDO教程。

查询可能会由于各种原因而失败，在这种情况下，mysql_*和mysqli扩展都会从各自的查询函数/方法中返回false。您需要测试该错误条件并相应地处理它。

mysql_扩展:

mysql_函数已弃用，在php版本7中已被删除。

在将$result传递给mysql_fetch_array之前检查$result。您会发现它是假的，因为查询失败了。请参阅[mysql_query][1]文档了解可能的返回值以及如何处理它们的建议。

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

if($result === FALSE) { 
    trigger_error(mysql_error(), E_USER_ERROR);
}

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

试试这段代码，它工作得很好

将post变量赋给变量

   $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'];
     }
}

如果数据库未选中，请检查，因为有时数据库未选中

检查

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

MySQL查询前 然后进入下一步

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

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

因为$username是一个PHP变量，我们需要将它作为字符串传递给mysqli，所以在查询中，你以单引号开始，我们将使用双引号，单引号和句号来连接("'.$username.'")，如果你以双引号开始，你将反转引号('".$username."')。

$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'];
     }

$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'];
     }

但是Mysql的使用已经贬值了很多，改用PDO。它很简单，但非常安全