使用PHP重新格式化JSON数组 - Reformat JSON array using PHP

Reformat JSON array using PHP

本文关键字：JSON 数组格式化 PHP 使用 | 更新日期: 2023-09-27

我有一个小的PHP脚本，它拉入一个CSV文件并从中创建一个JSON数组。

然而，我想改变我输出的JSON数组的格式。

PHP:

<?php
$file = fopen('food.csv', 'r');
$allfile = [];
$idsColumnsWanted = array_flip([0, 1, 2]);
while (false !== $fields = fgetcsv($file)) {
    $allfile[] = array_intersect_key($fields, $idsColumnsWanted);
}
fclose($file);
?>

输出:

var data = [["McDonalds","Fast Food","London"],["Marios","Italian","Manchester"]];

如何将CSV转换为以下格式:

var data = [
{name:"McDonald's Fast Food",location:"London"},
{name:"Marios Italian",location:"Manchester"}
];

它合并了前两项然后加上name &location .

我的food.csv文件是:

McDonalds,Fast Food,London
Marios,Italian,Manchester

下次请自己尝试:

$data = [["McDonalds", "Fast Food", "London"], ["Marios", "Italian", "Manchester"]];
$newData = [];
foreach ($data as $info) {
    $newData[] = [
        'name' => $info[0] . " " . $info[1],
        'location' => $info[2]
    ];
}
var_dump(json_encode($newData));

string '[{"name":"McDonalds Fast Food","location":"London"},{"name":"Marios Italian","location":"Manchester"}]' (length=104)

您需要创建一个所需格式的新数组，并简单地json_encode它。

如果你考虑使用jQuery -这就是

var data = [["McDonalds","Fast Food","London"],["Marios","Italian","Manchester"]];
var formated_data = {};
$.each(data, function(key, val) {
    formated_data[key] = {name: val[0] + ' ' + val[1], location: val[2]}
});
console.log(formated_data);

打开CSV文件，然后将CSV解析为数组。顺行

    <?php 
     $file="1_23.csv";
     $csv= file_get_contents($file);
     $array = array_map("str_getcsv", explode("'n", $csv));
     $json = json_encode($array);
     print_r($json);
    ?>