Question

我有一个ajax函数，它将一些数据提交给我的PHP文件。在我的PHP文件中，我有一些检查，它与数据，我然后尝试发出警报，说出来。虽然我无法收到任何警报。所有发生的事情都在控制台 - ＆gt; network-＆gt; submit.php-＆gt;响应中显示应该发出警报的javascript。

这是我的Ajax，我知道它有效：

$('#submit').click(function() {
    if(selectedImageArray.length <= 10){
  $.ajax({
    method: "POST",
    url: "submit.php",
    data: {selectedImageArray: selectedImageArray}
  }).done(function( msg ) {
    //alert( "Data Saved: " + selectedImageArray );
  });   }else{
        alert( "You can only trade up to 10 items at once." );
    }
});

然后在submit.php中我做了一些检查，并且有这个：

        if($QueueCount == 1){
            Echo '<script language="javascript">';
            Echo 'alert( "You already have an active trade, check your Steam profile.");';
            echo '</script>';
        }else{
            Echo '<script language="javascript">';
            Echo 'alert("You are currently in the trade Queue, you should receive a trade request soon.");';
            echo '</script>';
        }

这应该发出提醒，但它只是将<script language="javascript">alert( "You already have an active trade, check your Steam profile.");</script>输出到网络标签中，就像我说的那样。

如何使这项工作从PHP文件中弹出警报？

Answer 1

你试图从php输出javascript而不在你的ajax回调中输出它，将无法正常工作。

做你想做的最简单的方法是用你的php返回字符串，然后处理输出到你的javascript。

类似的东西：

if($QueueCount == 1){
        Echo "You already have an active trade, check your Steam profile.";
    }else{
        Echo "You are currently in the trade Queue, you should receive a trade request soon.";
    }

和javascript

$('#submit').click(function() {
  if(selectedImageArray.length <= 10){
      $.ajax({
        method: "POST",
        url: "submit.php",
        data: {selectedImageArray: selectedImageArray}
      }).done(function( msg ) {
        alert(msg);
      });   
  }else{
        alert( "You can only trade up to 10 items at once." );
    }
});

Answer 2

你的javascript代码必须如下所示：

$('#submit').click(function() {  
  if(selectedImageArray.length <= 10){
    $.ajax({
      method: "POST",
      url: "submit.php",
      data: {selectedImageArray: selectedImageArray},
      success: function(data,status){
        eval(data)
      }
    });   
  }else{
    alert( "You can only trade up to 10 items at once." );
  }
});

你的php代码不应该声明脚本标签：

    if($QueueCount == 1){
        echo 'alert( "You already have an active trade, check your Steam profile.");';
    }else{
        echo 'alert("You are currently in the trade Queue, you should receive a trade request soon.");';
    }

确保这是php文件的唯一输出。

此代码在开始生产之前还需要进行一些改进，例如在服务器端处理时回调错误，检查ajax是否由于网络问题和XSS保护而完全失败，列举一些。

用PHP和ajax发出alert（）

2 个答案: