Question

我有一个页面，其中有表格，每行都有一个图标以获取详细信息。单击该图标时，我将该行的唯一标识符转换为datastring并发送AJAX请求。我收到了回复，但无法得到正确的回复。为什么我做错了？我需要访问home.php中的$ desc，如何在响应中捕获它？

home.php

$.ajax({
        type: "POST",
        url: "actionone.php",
        data: dataString,
        cache: true,  
        success:function() {
        alert("success"); 
    }
});

actionone.php

if(isset($_POST['content']))
{
$content=(int)$_POST['content'];
$fetch = mysql_query("SELECT ISSUEDESCRIPTION FROM issues WHERE KBID =$content");
while($rowe = mysql_fetch_array($fetch))    
{   
    $desc = $rowe['ISSUEDESCRIPTION'];
}

Answer 1

对于ajax成功的工作，你输出一些PHP代码即将PHP代码更改为

if(isset($_POST['content']))
{
    $content=(int)$_POST['content'];
    $fetch = mysql_query("SELECT ISSUEDESCRIPTION FROM issues WHERE KBID =$content");
    while($rowe = mysql_fetch_array($fetch))    
    {   
         $desc = $rowe['ISSUEDESCRIPTION'];
    }
    echo $desc;
}

<强> JQUERY

$.ajax({
        type: "POST",
        url: "actionone.php",
        data: dataString,
        cache: true,  
        success:function(data) {
        alert(data.trim()); 
    }
});

没有从php中的ajax请求收到响应

1 个答案: