Question

我将数据发送到另一个php文件，然后成功检索数据以显示在div中。但是当我加载我的data.php脚本时，通常会显示通知＆＃34;未定义的索引：状态在....＆＃34;代码上的那一行是$ var = $ _POST [＆＃39; status＆＃39;];。

这是重新发布/编辑，因为脚本现在实际上有效，但我不明白为什么通知仍在显示？！

＆＃13;

<html>
<head>


<script type="text/javascript" src="jquery.js"></script>

<script type="text/javascript">

//daddy code
$ (document).ready(function() {

//mama code
$("button#postbutton").click(function() {

 var data = $("#formpost").serialize();


$.ajax({

type: "POST",

url: "data.php",

data: data,

success: function(data) {

$("#statustext").html(data);

}

});




});



});









</script>
</head>

<body>







<div id="global">

<form id="formpost" action="" method="post" onsubmit="return false">

<textarea id="text1" name="status" ></textarea>


<button id="postbutton">POST</button>

<a href="logout.php">LOGOUT</a>

</form>

<br/>
<br/>


<div id="allstatus">



<!-- SKELETON -->


<div id="wholestatus">


<div id="statuspic">
</div>


<div id="statusinfo">

<div id="statusname">JOnathan</div>
<div id="statustext"> </div>
<div id="statusoption"><button id="likestatus">LIKE</button></div>
<div id="statusoption"><button id="commentstatus">COMMENT</button></div>
<div id="statusoption"><button id="sharestatus">SHARE</button></div>
<div id="statusoption"><button id="statustime">TIME</button></div>






</div>




</div>	




<!-- SKELETON -->








</div>




</div>







</body>

＆＃13;

我的数据文件data.php

＆＃13;

<?php



$var = $_POST['status'];

const DB_HOST = 'localhost';
const DB_USER = 'root';
const DB_PASS = '';
const DB_NAME = 'forum';
//connecting 

$conn = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);

if($conn->connect_error) {
	die("Connection Failed: " . $conn->connect_error);
} else {



}


$sql = "INSERT INTO `question`(id, question) VALUES ('', '{$var}')";

$result = $conn->query($sql);

if($result) {

	
}
 else {

 	echo "failed: " . $conn->error;
 }


echo  "{$var}" ;


?>

＆＃13;

Answer 1

添加以下行/检查

$var = '';
if(isset($_POST['status'])){
  $var = $_POST['status'];
}

Answer 2

试试这个，

<?php

const DB_HOST = 'localhost';
const DB_USER = 'root';
const DB_PASS = '';
const DB_NAME = 'forum';
//connecting 

$conn = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);

if($conn->connect_error) {
die("Connection Failed: " . $conn->connect_error);
} else {

 }

if(isset($_POST['status']))
{
$var = $_POST['status'];

$sql = "INSERT INTO `question`(id, question) VALUES ('', '{$var}')";
$result = $conn->query($sql);
if($result) {

    }
    else {

    echo "failed: " . $conn->error;
   }
echo  "{$var}" ;
}

 ?>

JQuery / AJAX脚本有效，但仍然显示通知？

2 个答案: