如何使用php查找json文件中的数据 - How to lookup data in a json file using php

使用PHP，我希望通过url传递一个id，然后在JSON文件中查找一些数据…然后在页面上显示数据。

我将url设置为http://mytestapp.php?id=12345678，然后使用;

$id = $_GET['id'];

设置id变量。然后我有一个JSON如下;

{
    "ads": 
    [
        {   "id":"12345678", 
            "hirername":"Test Hirer", 
            "hirercontact":"Rob A",
            "role":"Ultra Sat Role",
            "requirements": [
                {"req":"Right to work in Australia"},
                {"req":"Live locally"}],
            "candidates": [
                {"name":"John Smith","dist":"Nunawading (23km away)","exp1":"Pizza maker at Don Domenicos for 4 years","exp2":"Bakery Assistant at Woolworths for 4 years","req":"","avail1":"Mon to Fri | Morning, Evening & Night","avail2":"","call":"0413451007"},
                {"name":"Jack Smith","dist":"Endeadvour Hills (35km away)","exp1":"Pizzaiolo (Pizza maker) at Cuor Di Pizza for 1 year","exp2":"","req":"","avail1":"Mon to Fri | Morning & Evening","avail2":"","call":"041345690"}]
        }, 
        {   "id":"12345679", 
            "hirername":"Test Hirer 2", 
            "hirercontact":"Jesse S",
            "role":"Ultra Sat Role 2",
            "requirements": [
                {"req":"Right to work in Australia"},
                {"req":"Live locally"}],
            "candidates": [
                {"name":"Jill Smith","dist":"Nunawading (23km away)","exp1":"Pizza maker at Don Domenicos for 4 years","exp2":"Bakery Assistant at Woolworths for 4 years","req":"","avail1":"Mon to Fri | Morning, Evening & Night","avail2":"","call":"0413451007"},
                {"name":"Jenny Smith","dist":"Endeadvour Hills (35km away)","exp1":"Pizzaiolo (Pizza maker) at Cuor Di Pizza for 1 year","exp2":"","req":"","avail1":"Mon to Fri | Morning & Evening","avail2":"","call":"041345690"}]
        }
    ]
}

我想搜索id，然后能够回显数据的内容。

我正在读取JSON并解码为数组;

    $json = file_get_contents('data.json');
    $arr = json_decode($json, true);

但我不知道现在如何读取数组，找到我想要的数据基于id，然后拉出数据，这样我就可以在页面上显示它如下;

Hirer: Test Hirer
Contact: Rob A 
Role: Ultra Sat Role

Requirements: 
- Right to work in Australia
- Live Locally

John Smith Nunawading (23km away) 
Pizza maker at Don Domenicos for 4 years 
Bakery Assistant at Woolworths for 4 years 
Mon to Fri | Morning, Evening & Night 
0413451007

Jack Smith Endeadvour Hills (35km away) 
Pizzaiolo (Pizza maker) at Cuor Di Pizza for 1 year 
Mon to Fri | Morning & Evening 
041345690

任何想法?

感谢Rob .

从@RobbieAverill借来的例子和修改，以满足您的需求，请检查这是否有效。

<?php
    $id = $_GET['id']; 
    $json = file_get_contents('data.json');
    $foundAd = null;
    $json = json_decode($json,true);
    foreach ($json['ads'] as $ad) {
        if ($ad['id'] == $id) {
            $foundAd = $ad;
            break;
        }
    }
    echo "Hirer:".$foundAd['hirername']."<br/>";
    echo "contact:".$foundAd['hirercontact']."<br/>";
    echo "role:".$foundAd['role']."<br/><br/>";
    echo "Requirements<br/>";
    echo "<ul>";
    foreach($foundAd['requirements'] as $req){
        echo "<li>".$req['req']."</li>";            
    }
    echo "</ul><br/>";
    foreach($foundAd['candidates'] as $req){
        echo $req['name']." ". $req['dist']."</br>";
        echo $req['exp1']."</br>";
        echo $req['exp1']."</br>";
        echo $req['avail1']."</br>";
        if($req['avail2']!=""){
            echo $req['avail2']."</br>";;       
        }
        echo $req['call']."</br></br>";
    }
?>

在当前的实现中，您需要循环遍历所有广告对象，例如

foreach ($arr['ads'] as $ad){
    if ($ad['id'] == $id){
        //do stuff;
    }
}

更好的实现是在存储json时使用id的值作为json对象的键。使用像

这样的东西

$ads[id] = $yourjsonobject;

那么引用就是$arr['ads'][id];。

你可以使用多个foreach或者如果你的键是已知的，只使用键来输出你需要的对象，如

echo $ad["hirername"];

使用foreach循环打印完整的对象:

foreach( $ad as $value){
    print_r($value);
}