jQuery.post没有从php脚本接收错误

时间:2013-05-17 07:59:54

标签: php javascript jquery post

我知道以前曾经问过这个问题,而且我已经查看了我能找到的每个帖子。我仍然无法获得jQuery.post函数来正确地从PHP脚本接收错误。这两个都是。
PHP:

<?php 
##CONFIGURATION
# LIST EMAIL ADDRESS
$toemail = "email here";

# SUBJECT (Subscribe/Remove)
$subject = "Someone has contacted International Designs";

# RESULT PAGE
$location = "../thank-you.php";

## FORM VALUES ##
$myname = $_REQUEST['myname'];
$myemail = $_REQUEST['myemail'];
$mymessage = $_REQUEST['mymessage'];

if ( empty($myname) || empty($myemail) || empty($mymessage) ) {

    exit('{"error":"That value was invalid!"}')

} else {

    # SENDER
    $email = $myname . " <" . $myemail . ">";

    # MAIL BODY
    $body .= "Name: " . $myname . " \n\n";
    $body .= "Email: " . $myemail . " \n\n";
    $body .= "Message: " . $mymessage . " \n\n";
    # add more fields here if required

    ## SEND MESSGAE ##

    mail( $toemail, $subject, $body, "From: $email" ) or die ("Mail could not be sent.");

}

?>

JS:

if (verify(myname, myemail, mymessage, human, hash, patt)) {
  $.post(myform.attr('action'), myform.serialize(), function() {
    $('#email-success').fadeIn();
    myform[0].reset();
    setTimeout("$('#email-success').fadeOut();", 5000);
  }, 'json')
  .fail(function() {
    alert('An error has occurred. Please try again later.')
  });
}

我已经尝试过5种不同的方法,但都没有。当我把&#39; json&#39;作为.post函数中的数据类型,.fail 总是触发,无论php脚本中有什么内容。如果我保留数据类型,那么.fail 永远不会在任何情况下触发。我有一种感觉问题是php命令和数据类型。任何帮助表示赞赏。

3 个答案:

答案 0 :(得分:2)

也许是因为您没有更改回复的http标头代码,也没有指定您的数据类型响应。

你使用&#34; 200来对前端(jQuery)代码进行php代码响应 - OK&#34;在任何情况下都是状态,但是在某些情况下,您会发现错误,其中包含&#39; HTTP / 1.0 400 - 错误请求&#39;响应或“HTTP / 1.0 500内部服务器错误&#39;。

而且,像Eggplant一样,您必须将响应数据类型指定为&#39; Content-Type:application / json&#39;。

所以,最终的代码是这样的:

<?php 
...

header('HTTP/1.0 204 – No Content', true, 204);
header('Content-Type: application/json');

if ( empty($myname) || empty($myemail) || empty($mymessage) ) {

    header('HTTP/1.0 400 – Bad Request', true, 400);
    exit('{"error":"That value was invalid!"}')

} else {

    ...

    $send = mail( $toemail, $subject, $body, "From: $email" );
    if (!$send)
        header('HTTP/1.0 500 – Internal Server Error', true, 500);
        exit ('{"error":"Mail could not be sent."}');
    }
}
return;
?>

对于警告Cannot modify header information - headers already sent by (output started at /homepages/.../design/contact.php:1),您可以检查同一问题的this answer

输出可以是:

  • 非故意:
    • 之前的空白
    • UTF-8字节顺序标记
    • 以前的错误消息或通知
  • 故意:
    • print,echo和其他产生输出的函数(如var_dump)
    • 之前的原始区域

聊天会话后更新:它是UTF-8 BOM问题

答案 1 :(得分:0)

虽然你的是一个有效的JSON数据,我总是建议使用json_encode()函数来创建JSON字符串。

exit(json_encode(array("error" => "That value was invalid!")));

另一个是确保发送正确的标头以确保脚本知道它的json数据。所以

header('Content-Type: application/json');
exit(json_encode("error" => "That value was invalid!"));

答案 2 :(得分:0)

试试这个

if (verify(myname, myemail, mymessage, human, hash, patt)) {
    $.ajax({
        type    :'POST',
        data    :$("#myformid").serialize(),
        beforeSend:function(){

        },
        success:function(res){
            if(res=='ok'){
                setTimeout("$('#email-success').fadeOut();", 5000);
            }else{
                //read respon from server
                alert(res)
            }
        },
        error:function(){
            //error handler
        }
    });
}

只是示例

$send = mail(....);//your mail function here
echo ($send) ? "ok" : "Error";