Question

我为生成器创建了一个名为look_api.php的脚本。

但是当我尝试它时，我得到了这个：

Parse error: syntax error, unexpected ';' in C:\xampp\htdocs\look_api.php on line 3

页面：

<?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;在线我得到这个错误：

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>

那么......我该如何解决？

Answer 1

你错过了)：

 $user = str_replace("'", "\\\'", str_replace('"', '\\"', $_GET['user']);

应该是：

 $user = str_replace("'", "\\\'", str_replace('"', '\\"', $_GET['user']));

你也有错误：

if($username = NULL) { $username = "Yvan" };

应该是：

if($username == NULL) { $username = "Yvan"; };

Answer 2

第3行和第4行出错

你遗失的第3行)
第4行分号应在if语句

之内

替换它..

$user = str_replace("'", "\\\'", str_replace('"', '\\"', $_GET['user'])); 
if($username == NULL) { $username = "Yvan"; }

Answer 3

替换

if($username = NULL) { $username = "Yvan" };

使用：

if($username = NULL) { $username = "Yvan"; }

解析错误：语法错误，意外';'在第3行的C：\ xampp \ htdocs \ look_api.php中

3 个答案: