我得到了一个名为look_api.php的生成器脚本。
但是当我尝试它时，我得到了：
Parse error: syntax error, unexpected ';' in C:\xampp\htdocs\look_api.php on line 3
这一页：
<?php
error_reporting(1);
$user = str_replace("'", "\\\'", str_replace('"', '\\"', $_GET['user']);
if($username = NULL) { $username = "Yvan" };
$con=mysqli_connect("MYSQL_IP","USER","PASSWORD","DATABASE");
// Mysql Connection
if (mysqli_connect_errno())
{
echo "Ai, check je mysql connection!: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT look FROM users WHERE username = '".$username."'");
while($row = mysqli_fetch_array($result))
{
?>
<?php
}
mysqli_close($con);
?>
</div> </div>
当我删除; 在行上我得到这个错误：
Parse error: syntax error, unexpected T_IF in C:\xampp\htdocs\look_api.php on line 4
脚本中的代码就在那一刻：
<?php
error_reporting(1);
$user = str_replace("'", "\\\'", str_replace('"', '\\"', $_GET['user'])
if($username = NULL) { $username = "Yvan" };
$con=mysqli_connect("MYSQL_IP","USER","PASSWORD","DATABASE");
// Mysql Connection
if (mysqli_connect_errno())
{
echo "Ai, check je mysql connection!: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT look FROM users WHERE username = '".$username."'");
while($row = mysqli_fetch_array($result))
{
?>
<?php
}
mysqli_close($con);
?>
</div> </div>
所以...我该如何解决？
您错过)了这里：
$user = str_replace("'", "\\\'", str_replace('"', '\\"', $_GET['user']);
应该 ：
$user = str_replace("'", "\\\'", str_replace('"', '\\"', $_GET['user']));
您还存在以下错误：
if($username = NULL) { $username = "Yvan" };
应该 ：
if($username == NULL) { $username = "Yvan"; };
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