Question

我有一个小的PHP脚本，可以输入CSV文件并从中创建一个JSON数组。

但是，我想更改输出的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转换为以下内容：

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

所以它基本上合并了前两个项目并添加了name＆amp; location。

我的food.csv文件是：

McDonalds,Fast Food,London
Marios,Italian,Manchester

Answer 1

下次请亲自尝试一下：

$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));

<强>输出

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

您需要以所需的格式创建一个新数组，并简单json_encode。

Answer 2

如果你考虑使用jQuery - 那就是

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);

Answer 3

打开CSV文件，然后将CSV解析为数组。流这一行，

    <?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);
    ?>

使用PHP重新格式化JSON数组

3 个答案: