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
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 $row['FirstName'];
}
}
其他回答
在执行取回数组之前，你也可以检查$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'];
}
}
试试下面的代码。它可能会工作得很好。
$username = $_POST['username'];
$password = $_POST['password'];
$result = mysql_query("SELECT * FROM Users WHERE UserName ='$username'");
while($row = mysql_fetch_array($result))
{
echo $row['FirstName'];
}
试试这个，它必须工作，否则您需要打印错误来指定您的问题
$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'];
}
<?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'];
}
?>
