Json.parse：意外的非空白字符？ [英] Json.parse: unexpected non-whitespace character?

问题描述

你好朋友，

我面对这个问题请帮帮我，我从php文件返回的数组我无法访问我的ajax代码中的元素，你可以帮我plz，

我的PHP代码



<？php

require_once（connection.php）;

global $ conn;



if（isset（$ _ POST ['expGoodsName']））{



$ expCheckSNo = $ _ POST ['expCheckSNo'];

$ goodsName = $ _ POST ['expGoodsName'];



$ goodsNamequery = mysqli_query（$ conn，SELECT S.Quantity，S.Unit，S.UnitPrice FROM tblstock S

INNER JOIN tblimportcheck imp USING（ImpSNo）

INNER JOIN tblexportcheck exp USING（ImpSNo）

WHERE exp.ExpSNO = $ expCheckSNo AND S.GoodsName ='$ goodsName'）; < br $>


$ rows = mysqli_fetch_assoc（$ goodsNamequery）;



$ qty = $ rows ['Quantity'] ;

$单位= $行['单位'];

$ UnitPrice = $ rows ['UnitPrice'];



$ myArr =数组（$ qty，$ Unit，$ UnitPrice）;



$ myJSON = json_encode（$ myArr）;



echo $ myJSON;



}

？>





这里是ajax代码：

$（＃expGoodsName）。change（函数getQuantity（参数）{

sendImportCheck（）;

}）;



函数sendImportCheck（）

{

var XMLHttp;

if（window.XMLHttpRequest）{

//现代浏览器的代码

XMLHttp = new XMLHttpRequest（）;

} else {

//旧IE浏览器的代码

XMLHttp = new ActiveXObject（Microsoft.XMLHTTP）;

}



var expCheckSNo = $（＃expCheckSNo）。val（）;

var expGoodsName = $（＃expGoodsName）。val（）;



XMLHttp.onreadystatechange = function（）

{

if（XMLHttp.status == 200& & XMLHttp.readyState == 4）{



var myresult = JSON.parse（XMLHttp.responseText）;

alert（myresult [0]）; //它应显示第一个元素



}

}

XMLHttp.open（POST，ExpCheckEdit.php，true）;

XMLHttp.setRequestHeader（Content-type，application / x-www-form -urlencoded）;

XMLHttp.send（expCheckSNo =+ expCheckSNo +& expGoodsName =+ expGoodsName）;

}



我尝试了什么：



我在谷歌尝试了很多但没有改变

Hello friends,
I face to this problem please help me, the array which I return from php file I can not access its element in my ajax code, can you help me plz,
its my php code

<?php
require_once("connection.php");
global $conn;

if (isset($_POST['expGoodsName'])) {

$expCheckSNo=$_POST['expCheckSNo'];
$goodsName=$_POST['expGoodsName'];

$goodsNamequery= mysqli_query($conn,"SELECT S.Quantity, S.Unit, S.UnitPrice FROM tblstock S
INNER JOIN tblimportcheck imp USING(ImpSNo)
INNER JOIN tblexportcheck exp USING(ImpSNo)
WHERE exp.ExpSNO =$expCheckSNo AND S.GoodsName='$goodsName' ");

$rows=mysqli_fetch_assoc($goodsNamequery);

$qty=$rows['Quantity'];
$Unit=$rows['Unit'];
$UnitPrice=$rows['UnitPrice'];

$myArr = array($qty,$Unit, $UnitPrice);

$myJSON = json_encode($myArr);

echo $myJSON;

}
?>


here is ajax code:
$("#expGoodsName").change(function getQuantity(argument) {
sendImportCheck();
});

function sendImportCheck()
{
var XMLHttp;
if (window.XMLHttpRequest) {
// code for modern browsers
XMLHttp = new XMLHttpRequest();
}else {
// code for old IE browsers
XMLHttp = new ActiveXObject("Microsoft.XMLHTTP");
}

var expCheckSNo=$("#expCheckSNo").val();
var expGoodsName=$("#expGoodsName").val();

XMLHttp.onreadystatechange=function()
{
if(XMLHttp.status==200 && XMLHttp.readyState==4) {

var myresult= JSON.parse(XMLHttp.responseText);
alert(myresult[0]);//it should display first element

}
}
XMLHttp.open("POST","ExpCheckEdit.php",true);
XMLHttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
XMLHttp.send("expCheckSNo="+expCheckSNo+"&expGoodsName="+expGoodsName);
}

What I have tried:

I tried a lot in google but nothing changed

推荐答案

conn;



if（isset（

conn;

if (isset(

_POST ['expGoodsName' ]））{

_POST['expGoodsName'])) {

expCheckSNo =

expCheckSNo=

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