SyntaxError：意外的JSON输入结束 [英] SyntaxError: Unexpected end of JSON input

问题描述

我可以在我的db中记录数据，但是ajax加载并不介入成功方法。问题是什么？

  error：SyntaxError：JSON输入的意外结束
 Object.parse（native）
 at n.parseJSON
  

record_data.php

 <？php 
 $ type = getPar_empty（type，$ _ GET，）; 
 $ new_nome = $ _ GET ['new_nome']; 
 $ new_cognome = $ _ GET ['new_cognome']; 
 $ new_email = $ _ GET ['new_email']; 
 $ new_lingua = $ _ GET ['new_lingua']; 
 if（$ type ==add_user）{
 $ stmt = null; （'$ new $''，'$ new_nome'，'$ new $'，'$'$'$'$' $ new_cognome， '$ new_lingua'， '手动'，现在（），0））; 
 if（！$ stmt）{
 log_msg（$ db-> error）; 
 die（）; 
} 
 $ stmt-> execute（）; 
 $ stmt-> close（）; 
 $ db-> close（）; 
} 
？> 
  

脚本
< $（$＃$）$（$＃$）$（$＃$）$（$＃$）$ = $（'＃cognome'）。val（）;
var email = $（'＃email'）。val（）;
var lingua = $（'＃lingua'）。val（） ;
var dataString ='nome ='+ nome +'& cognome ='+ cognome +'& email ='+ email +'& lingua ='+ lingua +'& type = add_user';
$ .ajax（{
类型：'GET'，
data：dataString，
url：record_data.php，
成功：函数（结果）{
$（＃status_text）。html（result）;
$（'＃nome'）。val（''）;
$（'＃cognome'）。val（''）;
$（'＃email'）。val（''）;
}，
错误：function（xhr，status，error）{
console.log（error）;
}
}）;
}）;

$（'＃adduser'）。submit（function（）{
return false;
}）;

表格

 < form name =adduserid =addusermethod =GETaction =＃> 
< div class =col-md-3> 
< div class =form-group m-b-30> 
< p>电子邮件< / p> 
< input class =form-controltype =emailid =emailname =emailplaceholder =indirizzo e-mailemail required> 
< / div> 
< / div> 
< div class =col-md-3> 
< div class =form-group m-b-30> 
< p> Nome< / p> 
< input class =form-controltype =textid =nomename =nomeplaceholder =nome> 
< / div> 
< / div> 
< div class =col-md-3> 
< div class =form-group m-b-30> 
< p> Cognome< / p> 
< input class =form-controltype =textid =cognomename =cognomeplaceholder =cognome> 
< / div> 
< / div> 
< div class =col-md-3> 
< div class =form-group m-b-30> 
< p> Lingua< / p> 
< select class =form-controlid =linguaname =lingua> 
< option value =it> IT< / option> 
< option value =en> EN< / option> 
< / select> 
< / div> 
< input type =submitclass =btn btn-embossed btn-primary m-r-20id =salvaBtnvalue =Aggiungi utente>< / input> 
< div id =status_text/>< / div> 
< / div> 
< / form> 
  

解决方案

您的jquery ajax调用应该包含

dataType：'html'，

明确声明数据从服务器返回的是 html

实际上您正在使用 result 和将它直接放入ID为 statusText

---

的DOM节点中您必须解决一个主要问题：您的PHP脚本未返回（ echo ing）任何数据！请注意：真的应该使用 mysqli 。 / strong>为您的数据库查询 http://php.net/manual/en/book。 mysqli.php ，并使用预处理语句来建立查询（而不是通过连接字符串来建立SQL查询）。

I can record data in my db but ajax loading doesn't inter in success method. What's the problem?

error: "SyntaxError: Unexpected end of JSON input
    at Object.parse (native)
    at n.parseJSON "

record_data.php

<?php
$type=getPar_empty("type",$_GET,"");
$new_nome=$_GET['new_nome'];
$new_cognome=$_GET['new_cognome'];
$new_email=$_GET['new_email'];
$new_lingua=$_GET['new_lingua'];
if ($type=="add_user"){
  $stmt=null;
  $stmt=$db->prepare("INSERT INTO newsletter_utenti(email,abilitato,nome,cognome,lingua,lista,data_creazione,unsubscribe) VALUES('$new_email',1,'$new_nome','$new_cognome','$new_lingua','manual',now(),0)");
  if (!$stmt) {
    log_msg($db->error);
    die();
  }
  $stmt->execute();
  $stmt->close();
  $db->close();
}
?>

script

   $("#salvaBtn").click(function(){
  var nome = $('#nome').val();
  var cognome = $('#cognome').val();
  var email = $('#email').val();
  var lingua = $('#lingua').val();
  var dataString = 'nome='+nome+'&cognome='+cognome+'&email='+email+'&lingua='+lingua+'&type=add_user';
  $.ajax({
    type:'GET',
    data:dataString,
    url:"record_data.php",
    success:function(result) {
              $("#status_text").html(result);
              $('#nome').val('');
              $('#cognome').val('');
              $('#email').val('');
    },
    error:function(xhr,status,error) {
      console.log(error);
    }
  });
});

$('#adduser').submit(function (){
  return false;
});

form

<form name="adduser" id="adduser" method="GET" action="#">
            <div class="col-md-3">
              <div class="form-group m-b-30">
                <p>E-mail</p>
                <input class="form-control" type="email" id="email" name="email" placeholder="indirizzo e-mail" email required>
              </div>
            </div>
            <div class="col-md-3">
              <div class="form-group m-b-30">
                <p>Nome</p>
                <input class="form-control" type="text" id="nome" name="nome" placeholder="nome">
              </div>
            </div>
            <div class="col-md-3">
              <div class="form-group m-b-30">
                <p>Cognome</p>
                <input class="form-control" type="text" id="cognome" name="cognome" placeholder="cognome">
              </div>
            </div>
            <div class="col-md-3">
              <div class="form-group m-b-30">
                <p>Lingua</p>
                <select class="form-control" id="lingua" name="lingua">
                  <option value="it">IT</option>
                  <option value="en">EN</option>
                </select>
              </div>
              <input type="submit" class="btn btn-embossed btn-primary m-r-20" id="salvaBtn" value="Aggiungi utente"></input>
              <div id="status_text" /></div>
            </div>
          </form>

解决方案

your jquery ajax call should include

dataType: 'html',

to explicitly state that the data coming back from the server is html

in fact you're taking result and putting it straight into the DOM node with id statusText

---

Then you have to fix a major issue: your PHP script is not returning (echoing) any data !

---

Bottom note: you really should use mysqli for your db queries http://php.net/manual/en/book.mysqli.php and also use prepared statements to build up the query (instead building up the SQL query by concatenating strings).

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