use*_*506 3 javascript php ajax json
我知道以前曾有人问过这个问题,但我已经回顾了我可以找到的所有以前的帖子,但仍然无法使它正常工作。希望这个问题很简单,你们可以帮助我解决。
我无法从PHP脚本获取返回数组,并将其进行json_encoded转换为Javascript数组。如果删除dataType应该为json的条件,则成功函数将抱怨意外数据。
我有一个Javascript函数,该函数使用POST方法调用PHP脚本。在PHP脚本中,我将一些调试消息写入日志文件,并显示json_encoded数组的内容。当我检查使用JSONLint的内容是否符合JSON要求,但我的JavaScript函数始终出错。
PHP函数如下所示:
<?php
// Include the php logger class so that we can write log messages to a log file
require_once('/myscripts/phplogger.php');
// Set up php log file
$log = new Logging();
$log->lfile('/mylogfiles/php-debug.log');
// Log status
$log->lwrite('Starting debug');
// ...
// (there's some more stuff here to get the data from a MySQL database)
// ...
// The following line when read in the log file shows a valid JSON array
$log->lwrite('array '.json_encode($dataVnvMethods));
$json_dataVnvMethods = json_encode($dataVnvMethods);
$log->lwrite('Ending debug');
?>
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Javascript函数如下所示:
function jsgetvnvfilters(projid,repid,weekid)
{
$.ajax(
{
url: './getvnvfilter.php',
data: {'projid':projid, 'weekid':weekid, 'repid':repid},
type: 'post',
dataType: 'json',
success: function()
{
alert('success');
var prevnvlist = '<?php echo $json_dataVnvMethods ?>';
var vnvlist = JSON.parse(prevnvlist);
for (var x = 1; x <= vnvlist.length; x++) {
var vnv = vnvlist[x]['VnVMethod'];
vnvSel.options[vnvSel.options.length] = new Option(vnv, vnv);
}
},
error: function()
{
alert('failure');
}
}
);
}
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您误解了javascript和php的关系;该php仅在页面加载时呈现一次,因此您无法echo在javascript success处理程序中返回php数据。
相反,您的php脚本输出的所有内容都将在javascript变量中可用:
success: function(vnvlist) {
// ^^^^^^^ this is what has been outputted by php, the
// json already parsed
// your data is available in `vnvlist`
// var prevnvlist = '<?php echo $json_dataVnvMethods ?>';
// with dataType='json' you don't need to parse anything
// as jQuery will parse it for you
// var vnvlist = JSON.parse(prevnvlist);
// so all you need is this...
alert('success');
for (var x = 1; x <= vnvlist.length; x++) {
var vnv = vnvlist[x]['VnVMethod'];
vnvSel.options[vnvSel.options.length] = new Option(vnv, vnv);
}
},
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并且您需要确保php仅输出您的json字符串:
...
// The following line when read in the log file shows a valid JSON array
$log->lwrite('array '.json_encode($dataVnvMethods));
// output your data so that it is available on the client-side
echo json_encode($dataVnvMethods);
$log->lwrite('Ending debug');
...
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