[英]How to convert JSON to array and loop over it in jQuery?
我正在使用JSON與用戶進行通信。 PHP將數組轉換為JSON到這種形式:
{"success":"text-to-display","warning":"NONE","notice":"text-to-display","error":"NONE"}
jQuery顯示通知:
function callback (data){
if(data.notice !== 'NONE'){
displayNotice(data.notice);
}
if(data.success !== 'NONE'){
displaySuccess(data.success);
}
if(data.warning !== 'NONE'){
displayWarning(data.warning);
}
if(data.error !== 'NONE'){
displayError(data.error);
}
}
不幸的是,在這個方法中不能顯示兩個錯誤或兩個通知或兩個警告,因為新語句替換舊語句。
<?php
$uwaga['error'] = 'old statement';
$uwaga['error'] = 'new statement';
// display only "new statement"
echo json_encode($uwaga);
?>
我認為使用數組:
<?php
$uwaga = array();
$uwaga[1] = array('type' => 'notice', 'text' => 'old statement');
$uwaga[2] = array('type' => 'notice', 'text' => 'new statement');
// display "new statement" and "old statement"
// generate: {"1":{"type":"notice","text":"old statement"},"2": {"type":"notice","text":"new statement"}}
echo json_encode($uwaga);
?>
如何在jQuery上“翻譯”這個PHP代碼(主要是:如何將json對象轉換為數組?如何使用循環?如何使用此循環?如何引用$uwaga[$key]['name']
和$uwaga[$key]['text'])
?
foreach ($uwaga as $key => $value) {
switch ($uwaga[$key]['name']) {
case 'warning':
displayWarning($uwaga[$key]['text']);
break;
}}
好吧,假設我們有一個PHP數組
PHP:
<?php
$myArray = array(
"test1"=>array("name"=>"test1name", "value"=>"test1value"),
"test2"=>array("name"=>"test2name", "value"=>"test2value"),
"test3"=>array("name"=>"test3name", "value"=>"test3value")
);
// Now make a javascript variable containing echoed JSON
echo "<script type='text/javascript'>var returnedJSON = " . json_encode($myArray) . ";</script>";
這將輸出以下JSON,為您提供一個javascript對象:
var returnedJSON = {"test1":{"name":"test1name","value":"test1value"},"test2":{"name":"test2name","value":"test2value"},"test3":{"name":"test3name","value":"test3value"}};
使用Javascript:
//Once you have the variable from above which can come in various ways (through ajax, jsonp etc) you can iterate over it quite simply in jQuery
$.each(returnedJSON, function (index, value) {
console.log(index + ": " + value.name);
});
而不是這個:
$uwaga = array();
$uwaga[1] = array('type' => 'notice', 'text' => 'old statement');
$uwaga[2] = array('type' => 'notice', 'text' => 'new statement');
只需這樣做,沒有索引:
$uwaga = array();
$uwaga[] = array('type' => 'notice', 'text' => 'old statement');
$uwaga[] = array('type' => 'notice', 'text' => 'new statement');
這將在數組的末尾為它們分配索引(從零而不是一個)。
然后采取所有這些:
if(data.notice !== 'NONE'){
displayNotice(data.notice);
}
if(data.success !== 'NONE'){
displaySuccess(data.success);
}
if(data.warning !== 'NONE'){
displayWarning(data.warning);
}
if(data.error !== 'NONE'){
displayError(data.error);
}
...並將它們包裝在一個jQuery的each()
塊each()
如Rob推薦的那樣。 它將成為(假設數據在json.messages
):
$.each(json.messages, function (index, data) {
if(data.notice !== 'NONE'){
displayNotice(data.notice);
}
if(data.success !== 'NONE'){
displaySuccess(data.success);
}
if(data.warning !== 'NONE'){
displayWarning(data.warning);
}
if(data.error !== 'NONE'){
displayError(data.error);
}
});
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