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javascript - 将未定义分配给 JSON 属性值

问题描述

我无法从在 wordpress 环境中运行的服务器返回的 JSON 中提取 JSON 属性值。

客户端通用 Woody Snippet 脚本

<html>
<body>

<h2>JSON string output from a JavaScript object.</h2>

<h4>JSON String Value is:</h4><p id="demo"> </p>
<h4>JSON Symbol2 Value is:</h4><p id="json_symbol2"> </p>
<h4>JSON Symbol Value is:</h4><p id="json_symbol"> </p>
<h4>JSON Price Value is:</h4><p id="json_price"> </p>

<script>    
var ourRequest = new XMLHttpRequest();
ourRequest.open("GET", "https://dividendlook.co.uk/Editor-PHP-1.9.0/controllers/ajax_stock_holdings2.php"); 
ourRequest.onload = function() {
var obj =   ourRequest.responseText;
var myJSON =    JSON.parse(JSON.stringify(obj));
//var myJSONprice =     JSON.parse(JSON.stringify(obj.price));
console.log( myJSON );

console.log( myJSON.symbol );
console.log(myJSON.price);

document.getElementById("demo").innerHTML = myJSON;
//document.getElementById("json_symbol2").innerHTML = myJSONprice;
document.getElementById("json_symbol").innerHTML = myJSON.symbol;
document.getElementById("json_price").innerHTML = myJSON.price;
}
ourRequest.send();  
</script>
</body>
</html>

服务器 PHP 脚本

<?php

/*
 * ajax_stock_holdings2.php
 */
$filename = 'ajax_stock_holdings2.php';

 /* Loads the WordPress environment and template */
require( '../../wp-blog-header.php' );

global $current_user;
wp_get_current_user();

// DataTables PHP library
include( "../lib/DataTables.php" );


//$stock_id = $_GET['stock_id']; // manually setting variable for testing
$stock_id = 1293;
$stock_array = array();

// check if variable is NOT blank pass JSON back to client 
if ($stock_id  <> "") {

//echo "the value of stock_id is :" . $stock_id . ":" . "\n";

try {
    $pdo = new PDO(strtolower($sql_details['type']) . ":host=" . $sql_details['host'] . ";dbname=" . $sql_details['db'], $sql_details['user'], $sql_details['pass']);
    $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
//    echo $filename . "Connected successfully" . "\n\n";
    }
catch(PDOException $e)
    {
//    echo $filename . "Connection failed: " . $e->getMessage();
    }

$result = $pdo->query("SELECT id, symbol, name, price FROM dm_stocks WHERE id = $stock_id");
        foreach ($result as $row) {

        array_push( $stock_array, array('symbol'=>$row['symbol'], 'price'=>$row['price'] ) );
        }

echo json_encode($stock_array, JSON_PRETTY_PRINT | JSON_UNESCAPED_SLASHES | JSON_NUMERIC_CHECK); 
}

?>

返回的 JSON

[
    {
        "symbol": "GSK.LSE",
        "price": 1744
    }
]

客户端脚本的输出

Test JSON
JSON string output from a JavaScript object.
JSON String Value is:
[ { "symbol": "GSK.LSE", "price": 1744 } ]

JSON Symbol2 Value is:
JSON Symbol Value is:
undefined

JSON Price Value is:
undefined

我尝试取消注释 Symbol2 的注释行，这会导致错误，即在 stringify 之前分配属性值

VM23451:1 Uncaught SyntaxError: Unexpected token u in JSON at position 0
    at JSON.parse (<anonymous>)
    at XMLHttpRequest.ourRequest.onload ((index):303)

第 303 行是

var myJSONprice =   JSON.parse(JSON.stringify(obj.price));

非常感谢您对此问题的任何帮助，非常感谢 Colin

我正在尝试集成上述代码以将价格作为消息写入 datatables.net 编辑器模式，客户端片段提取如下

    editor.dependent( 'dm_transactions.stock_id', function ( val, data, callback ) {
    $.ajax( {
        url: '../../Editor-PHP-1.9.0/controllers/ajax_stock_transactions.php',
        dataType: 'json',
        // pass stock_id value to server php script
        data: { "stock_id": val },
        success: function (json) {          
            callback(json);

ourRequest = new XMLHttpRequest();
ourRequest.open("GET", "https://dividendlook.co.uk/Editor-PHP-1.9.0/controllers/ajax_stock_transactions.php");  
ourRequest.onload = function() {
obj =   ourRequest.responseText;    
myJSONobj =     JSON.parse(obj);
myJSONstr =    JSON.parse(JSON.stringify(obj));

console.log( myJSONobj );
console.log( myJSONstr );

console.log(myJSONobj[0].symbol);
console.log(myJSONobj[0].price);

document.getElementById("json_obj").innerHTML = myJSONobj;
document.getElementById("json_str").innerHTML = myJSONstr;
document.getElementById("json_symbol").innerHTML = myJSONobj[0].symbol;
document.getElementById("json_price").innerHTML = myJSONobj[0].price;
}
ourRequest.send();  

editor.field("dm_transactions.price").set(myJSONobj[0].price);
    }),{
        event: 'keyup change'
    };

    editor.dependent('dm_transactions.stock_id', function ( val, data, callback ){
        return { messages: { 'dm_transactions.price': 'Yesterday Close : ' + myJSONobj[0].price }};

    },{
    event: 'keyup change'
    });

产生的错误如下

(index):570 Uncaught TypeError: Cannot read property '0' of undefined
    at (index):570
    at HTMLDivElement.<anonymous> (dataTables.editor.js:2602)
    at HTMLDivElement.dispatch (jquery-3.3.1.js:5183)
    at HTMLDivElement.elemData.handle (jquery-3.3.1.js:4991)
    at Object.trigger (jquery-3.3.1.js:8249)
    at HTMLSelectElement.<anonymous> (jquery-3.3.1.js:8327)
    at Function.each (jquery-3.3.1.js:354)
    at jQuery.fn.init.each (jquery-3.3.1.js:189)
    at jQuery.fn.init.trigger (jquery-3.3.1.js:8326)
    at dataTables.editor.js:9159

线（索引）：570 低于

editor.field("dm_transactions.price").set(myJSONobj[0].price);

对于如何将此代码集成到上面显示的数据表代码中的任何帮助，将不胜感激，感谢 Colin

标签： javascriptphpjsonwordpress

解决方案

来自您的 PHP 代码的响应是一个关联数组的数组，它曾经json_encoded变成一个对象数组。因此，要访问您需要遍历数组的各个值，或者，如果查询只返回一行，您可以简单地使用[0]元素，例如

console.log(myJSON[0].symbol);
console.log(myJSON[0].price);

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