我曾在Stackoverflow中看到过与其相关的其他文章。试过这个代码,但它没有为我工作。使用AJAX检索JSON数据
我有一个代码,将从数据库中获取数据是这样的:
function getWrkNoTest($wrkno){
$conf = new BBAgentConf();
$log = new KLogger($conf->get_BBLogPath().$conf->get_BBDateLogFormat(), $conf->get_BBLogPriority());
$connection = MySQLConnection();
$getWrkNoTest ="";
$lArrayIndex = 0;
$query = mysql_query("SELECT a.jobinfoid, a.WRKNo, a.cate, a.det, a.compclosed, a.feedback, a.infoID, b.callerid, b.customername FROM
bb_jmsjobinfo a
LEFT JOIN bb_customer b ON a.customerid = b.customerid
WHERE a.WRKNo = '$wrkno';");
$result = mysql_query($query);
$log->LogDebug("Query[".$query."]");
while ($row = mysql_fetch_array($result)){
$getWrkNoTest = array("jobinfoid"=>$row['jobinfoid'],
"WRKNo"=>$row['WRKNo'],
"cate"=>$row['cate'],
"det"=>$row['det'],
"compclosed"=>$row['compclosed'],
"feedback"=>$row['feedback'],
"infoID"=>$row['customerid'],
"customerid"=>$row['infoID'],
"callerid"=>$row['callerid'],
"customername"=>$row['customername']);
$iList[$lArrayIndex] = $getWrkNoTest;
$lArrayIndex = $lArrayIndex + 1;
}
$QueryResult = print_r($getWrkNoTest,true);
$log->LogDebug("QueryResult[".$QueryResult."]");
closeDB($connection);
return $iList;
}
此代码将从数据库.Lets连接到MySQL数据库和检索数据说这个文件名是mysql.php。经过我有一个getjson.php文件命名,这个文件将从mysql.php获取数据并传送到阿贾克斯。这个文件的代码如下:
<?php
//getCustomerNames will display all customer name that use the same number phone ,displayed on *top callername slection
include 'mysql.php';
$wrkno = $_GET["wrkno"];
$jms = getWrkNoTest($wrkno);
foreach($jms as $findContact){
$findContact['jobinfoid'];
$findContact['WRKNo'] ;
$findContact['cate'] ;
$findContact['det'] ;
$findContact['compclosed'] ;
$findContact['feedback'] ;
$findContact['customerid'] ;
$findContact['infoID'];
$findContact['infoID'];
$findContact['customername'] ;
$matches[] = $findContact;
}
echo json_encode($matches);
?>
然后,这是我的Ajax代码,将分析所有数据到html:
$.getJSON("jsonJms.php", {wrkno: wrkno}, function(data) {
$.each(data, function(key,val) {
$("#cname").val(val.customername);
$("#cnumb").val(val.callerid);
$("#comp").val(val.compclosed);
$("#cate").val(val.cate);
$("#det").val(val.det);
$("#feed").val(val.feedback);
});
});
当这在浏览器中运行时,只是给出一个输出Null。请帮助我。任何帮助,我将在我的整个人生中感激。
如果你看看firebug中服务器的响应形式,实际上从jsonJms.php返回的是什么? – Jiminyjetson
@Jiminyjetson它没有给出任何回应 – art
好的,所以你的GET请求返回一个空的响应?如果你直接去[你的本地主机url] /jsonJMS.php?workno= [测试值],它是否仍然没有返回? – Jiminyjetson