json 字符串化为 php
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json stringify to php
提问by i need help
I want to pass the key values into php page.
我想将键值传递到 php 页面。
At php page, I will start to read value by matching ajaxcallid.
在 php 页面，我将通过匹配ajaxcallid.
But it not working.
但它不起作用。
It gotta do with syntax/way I am passing in causing error.
这与我传入的语法/方式有关，导致错误。
parse error
invalid json: ajax call id is missing
JavaScript/AJAX:
JavaScript/AJAX：
var person = {
"Address" : "123 Anywhere St.",
"City" : "Springfield",
"PostalCode" : 99999
};
alert(person);
person= JSON.stringify(person);
alert(person);
$.ajax({
url: 'ROOT_URL/admin/ajaxtest.php',
type: "POST",
dataType: 'json',
data: {ajaxcallid: '26', jsarr: person},
timeout: 5000,
success: function(output) {
alert(output.Address);
},
});
PHP:
PHP：
<?php
if (isset($_REQUEST['ajaxcallid']))
{
if($_REQUEST['ajaxcallid']==26)
{
//example, I want to read value of person.Address, person.City,
//person.PostalCode
//what is the easiest way
$phparr= json_decode($_REQUEST['jsarr']);
//do all other operation
$output= json_encode($phparr);
}
}
else
{
$output= "ajax call id is missing";
}
echo $output;
?>
回答by Evalds Urtans
If you are not using dataType : 'json', you might need to do stripslashes
如果你没有使用 dataType : 'json'，你可能需要做 stripslashes
$.post(window.data.baseUrl, {posts : JSON.stringify(posts)});
And in php:
在 php 中：
$posts = json_decode(stripslashes($_POST['posts']));
回答by amol challawar
This helped me:
这帮助了我：
data = json_decode($this->request->data['jsarr'], true);
in your php code for accessing the record
在用于访问记录的 php 代码中
Hope it will help someone!
希望它会帮助某人！
回答by muddybruin
I'm going to take a guess and say that you SHOULDN'T stringify anything. I believe JQuery will do that for you. Namely, no person = JSON.stringify(person). Give that a try.
我要猜测一下，你不应该把任何东西串起来。我相信 JQuery 会为你做到这一点。即，no person = JSON.stringify(person)。试一试吧。
回答by Peter Oram
This is what your $.ajaxcall and the PHPside should look like:
这就是您的$.ajax呼叫和PHP侧面应该是这样的：
JQuery
查询
$.ajax({
url: "/admin/ajaxtest.php",
method: "POST",
data: {
ajaxcallid: "26",
person: JSON.stringify({
"Address" : "123 Anywhere St.",
"City" : "Springfield",
"PostalCode" : "99999"
})
}
}).done(function(data) {
if (!data) {
// generic error message here
} else if (data == 'invalid') {
alert('no ajaxcallid received');
} else {
var result = $.parseJSON(data); // if you pass back the object
alert(result.Address);
}
});
PHP
PHP
if (isset($_REQUEST['ajaxcallid'])) {
if ((int) $_REQUEST['ajaxcallid'] == 26) {
$personData = json_decode($_REQUEST['person']);
$address = $personData->Address;
$postalCode = $personData->PostalCode;
$returnData = json_encode($personData);
echo $personData;
die();
}
} else {
echo 'invalid';
die();
}
回答by TheVillageIdiot
I have not worked with PHPbut from my experience with ASP.netfollowing may help you.
我没有工作过，PHP但根据我的经验，ASP.net以下可能会对您有所帮助。
Add contentTypekey to ajax settigns:
将contentType键添加到 ajax 设置：
type: "POST",
contentType:'application/json',
dataType: 'json',
also I think you need to stringifywhole value you are assigning to datalike this:
我也认为你需要像这样stringify分配给你的全部价值data：
var person = {
"Address" : "123 Anywhere St.",
"City" : "Springfield",
"PostalCode" : 99999
};
var d= {ajaxcallid: '26', jsarr: person};
var dat=JSON.stringify(d);
......
data: dat,
......
