Php 如何创建一个将显示json文件中多个元素的条件？

我想创建一个条件来实现这一点： 如果在一个句子中，部门名称等于json文件中的部门名称，我们将显示带有其他元素的句子，并与所选的部门名称链接，例如他的人口、他的部门代码

json文件看起来像这样，比这个长很多：

[
    {
        "datasetid": "population-francaise-par-departement-2018",
        "recordid": "7b81f8adc3cb71b942540e51d868992a7d588595",
        "fields": {
            "departement": "Orne",
            "code_departement": "61",
            "geom": {
                "type": "Polygon",
                "coordinates": [
                    [
                        -0.7399722943,
                        48.6217032013
                    ],
                    [
                        -0.7388681294,
                        48.622541151
                    ],
                    [
                        -0.4329546858,
                        48.8633021563
                    ],
                    [
                        -0.4305570769,
                        48.8630882975
                    ]
                ]
            },
            "geo_point_2d": [
                48.62307419424573,
                0.127896583868712
            ],
            "population": 282516
        },
        "geometry": {
            "type": "Point",
            "coordinates": [
                0.127896583868712,
                48.62307419424573
            ]
        },
        "record_timestamp": "2018-06-21T14:36:09.020+02:00"
    },
    {
        "datasetid": "population-francaise-par-departement-2018",
        "recordid": "7b81f8adc3cb71b942540e51d868992a7d588595",
        "fields": {
            "departement": "Blois",
            "code_departement": "28",
            "geom": {
                "type": "Polygon",
                "coordinates": [
                    [
                        -0.7399726385,
                        22.6217542752
                    ],
                    [
                        -0.7388681294,
                        55.622541151
                    ],
                    [
                        -0.4329546858,
                        47.8633021563
                    ],
                    [
                        -0.4305570769,
                        12.8630882975
                    ]
                ]
            },
            "geo_point_2d": [
                48.62307419424573,
                0.127896583868712
            ],
            "population": 254654
        },
        "geometry": {
            "type": "Point",
            "coordinates": [
                0.127896583868712,
                48.62307419424573
            ]
        },
        "record_timestamp": "2018-06-21T14:36:09.020+02:00"
    }
]

我试过这个：

<?php

$populationdata = file_get_contents('population-francaise-par-departement-2018.json');

$myfile = json_decode($populationdata, true);

foreach($myfile as $record) { 
$name = "Blois";
$sentence= 'you are in the department call '. $name . ' the population is about '. $thepopulation .' people ' . "the code  departement is: ". $codedepartement;

     if($record['fields']['departement'] === $nom){
         echo $sentence;

     }
}

这里的问题是： 代码显示句子，但代码部门的人口信息不是好的。我如何选择好的

提前谢谢你

每次在我的句子中，部门名称出现在json文件中时，我都希望这样显示：

---

您所在的部门呼叫人数为254654人，代码部门为：28

您可以尝试这样的函数。它在department字段中查找传递的部门名称，然后使用返回的索引查找人口和代码：

function show_population($data, $department) {
    $fields = array_column($data, 'fields');
    $k = array_search($department, array_column($fields, 'departement'));
    if ($k !== false) {
        $sentence = 'you are in the department call '. $department . ' the population is about '. $fields[$k]['population'] .' people ' . "the code  departement is: ". $fields[$k]['code_departement'] . PHP_EOL;
    }
    else {
        $sentence = 'the department ' . $department . ' has no data available' . PHP_EOL;
    }
    return $sentence;
}

echo show_population($myfile, 'Blois');
echo show_population($myfile, 'Paris');

输出：

you are in the department call Blois the population is about 254654 people the code  departement is: 28
the department Paris has no data available

---

在句子$thepopulation=$record['fields']['population']和$CodeDepartment=$record['fields']['code\u Department']之前定义此变量。您也没有定义$nom变量。定义i put$name=Blois这是什么意思？如果$record['fields'['Department']==$nom，请检查您的if条件{这是错误的变量$nom.oh好吧，我之所以这样做是因为我第一次尝试这样的东西$name=$record['fields']['department']；$thepopulation=$record['fields']['population']；$code department=$record['fields']['code_department']；在我的情况下写$name，但他没有工作