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Question

我的问题如下，我必须在3种情况下显示错误信息 1-缺少一些文字 2-电话号码应为所有数字 3-电子邮件格式无效 $ errmsg是一个JSON对象，应该在ajax中解析之后显示该消息。但是我不知道如何在ajax中解析它并返回值。这是我的代码：

$errmsg='{"invalid":"some text missing",
      "mailerr":"email format is incorrect",
      "telephoneerr":"telephone should be all digits"}';
 if(($name=="")||($email=="")||($telephone=="")||($username=="")||($password==""))
{
  echo $errmsg;
}elseif(!preg_match($pattern_email,$email)){
 echo $errmsg;
}
elseif (!preg_match($pattern_phone,$telephone)){
  echo $errmsg;
 }

和ajax代码如下：

  $.ajax({
          type:'POST',
          url: 'contactData.php',
          //dataType: "JSON",     
 data{"name":name,"telephone":telephone,"email":email,"username":username,"password":password},
       success: function(data) {            
      $("#validate").html(data);     
       }        
    });

Answer 1

您可以在ajax（在jQuery中）使用“dataType：JSON”而不是

$errmsg='{"invalid":"some text missing",
  "mailerr":"email format is incorrect",
  "telephoneerr":"telephone should be all digits"}';

在php

中返回一个带有json_encode的数组

echo json_encode(array("invalid" => "some text missing",
  "mailerr" => "email format is incorrect",
  "telephoneerr" => "telephone should be all digits"));

现在您可以像

这样的对象访问数组值

data.invalid OR data.mailerr

Answer 2

试试这个：

$errmsg='{"invalid":"some text missing",
         "mailerr":"email format is incorrect",
         "telephoneerr":"telephone should be all digits"}';

if (($name == "") || ($email == "") || ($telephone == "") || ($username == "") || ($password == "") || !preg_match($pattern_email, $email) || !preg_match($pattern_phone, $telephone)){
    return $errmsg;
}

更新

如果要使用jQuery解析JSON数据，可以查看以下代码：

<?php
if (($name == "") || ($email == "") || ($telephone == "") || ($username == "") || ($password == "")) {
    $errorMsg = array("error" => "some text missing");
} elseif (!preg_match($pattern_email, $email)) {
    $errorMsg = array("error" => "email format is incorrect");
} elseif (!preg_match($pattern_phone, $telephone)){
    $errorMsg = array("error" => "telephone should be all digits");
}

return json_encode($errorMsg);
?>
<script>
    $.ajax({
        type:'POST',
        url: 'contactData.php',
        dataType: "JSON",     
        data{"name":name, "telephone":telephone, "email":email, "username":username, "password":password},
        success: function(data) {
            data = jQuery.parseJSON(data);
            $("#validate").html(data.error);     
        }        
    });
</script>

Answer 3

看起来你的JSON形成得很糟糕。尝试在没有外部单引号的情况下编写它，如下所示：

$errmsg = {
    "invalid"      : "some text missing",
    "mailerr"      : "email format is incorrect",
    "telephoneerr" : "telephone should be all digits"
};

然后调用$errmsg['invalid']等来获取每个案例所需的错误消息。

Answer 4

我认为你的输出是正确的，你需要在jquery中显示

  $.ajax({
           type:'POST',
         url: 'contactData.php',

         //dataType: "JSON",
    data{"name":name,"telephone":telephone,"email":email,"username":username,"password":password},
   success: function(data) {

  var obj = jQuery.parseJSON(data);

  $("#validate").html(obj.invalid);



   }        
});

如何从ajax回显json语句

4 个答案:

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