我试图从一个MySQL表中选择数据，但我得到以下错误消息之一:
Mysql_fetch_array()期望参数1为给定的资源布尔值
这是我的代码:
$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'];
}
当前回答
$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'];
}
有时将查询抑制为@mysql_query(您的查询);
其他回答
$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'];
}
有时将查询抑制为@mysql_query(您的查询);
$result = mysql_query('SELECT * FROM Users WHERE UserName LIKE $username');
使用单引号定义字符串，PHP不解析单引号分隔的字符串。为了获得变量插值，您需要使用双引号或字符串连接(或两者的组合)。更多信息请参见http://php.net/manual/en/language.types.string.php。
此外，你应该检查mysql_query是否返回了一个有效的结果资源，否则fetch_*， num_rows等将不会对结果起作用，因为这不是一个结果!即:
$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 $row['username'];
}
}
http://us.php.net/manual/en/function.mysql-query.php获取更多信息。
在执行取回数组之前，你也可以检查$result是否失败
$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'];
}
}
试试这个
$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'];
}
}
<?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."'");
