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php - 如何在php中获取特定的数组值

转载 作者:可可西里 更新时间:2023-11-01 13:25:56 26 4
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我想在 php 中输出数组中的特定值

下面是我的代码,数组是$content

<?php
$content = $_POST;

for($i=1; $i < $content['itemCount'] + 1; $i++) {
$name = 'item_name_'.$i;
$quantity = 'item_quantity_'.$i;
$price = 'item_price_'.$i;
$image='item_image_'.$i;
$option='item_options_'.$i;
$total = $content[$quantity]*$content[$price];
}
?>

<?php
print_r( $content );

?>

输出如下:

Array ( [currency] => INR 
[shipping] => 0
[tax] => 0
[taxRate] => 0
[itemCount] => 3
[item_name_1] => Our Nest
[item_quantity_1] => 1
[item_price_1] => 1900
[item_options_1] => image: CF01108.jpg, productcode: 602793420
[item_name_2] => Our Nest
[item_quantity_2] => 1
[item_price_2] => 2100
[item_options_2] => image: CF01110.jpg, productcode: 123870196
[item_name_3] => Our Nest
[item_quantity_3] => 1
[item_price_3] => 1800
[item_options_3] => image: CF01106.jpg, productcode: 416267436 )

如何在 php 变量中获取 productcode 值并回显它?

例子:

602793420, 123870196, 416267436

最佳答案

你可以使用explode()函数得到productcode,像这样,

$product_code = explode("productcode: ", $option)[1];

引用如下:

所以你的代码应该是这样的:

<?php
$content = $_POST;

for($i=1; $i < $content['itemCount'] + 1; $i++) {
$name = 'item_name_'.$i;
$quantity = 'item_quantity_'.$i;
$price = 'item_price_'.$i;
$image='item_image_'.$i;
$option='item_options_'.$i;
$product_code = explode("productcode: ", $option)[1];
$total = $content[$quantity]*$content[$price];
}
?>

关于php - 如何在php中获取特定的数组值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34307906/

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