从JSON使用PHP提取数据 [英] Extracting Data from JSON Using PHP

问题描述

下面是一个测试杰森饲料

Here's a test jason feed

{"MEMBERS":[{"NAME":"Joe Bob","PET":["DOG"]}, {"NAME":"Jack Wu","PET":["CAT","DOG","FISH"]}, {"NAME":"Nancy Frank","PET":["FISH"]}]} 

我正在试图做的是提取的数据，如果PET含有猫或鱼或两者兼而有之。另一位用户提出了一个过滤器这样：

What I'm attempting to do is extract data if PET contains CAT or FISH or both. Another user suggested a filter as such:

$filter = array('CAT', 'FISH');
// curl gets the json data (this part works fine but is not shown for brevity)
$JSONarray=json_decode($JSONdata,true);
foreach($JSONarray['MEMBERS'] as $p) {
       if (in_array($p['PET'], $filter)) {
       echo $p['NAME'] . '</br>';
        }
}

但它不返回任何东西。

But it's not returning anything.

请注意：已编辑基于下面的评论

Note: edited based on comments below

推荐答案

它应该是这样的：

foreach($JSONarray['MEMBERS'] as $p) {

    if (array_diff($p['PET'], $filter) != $p['PET']) {
        echo $p['NAME'].'</br>';
    }
}

• 记住，总是使用引号当您试图访问一个关联数组的元素。不带引号，PHP会尝试跨preT它作为一个常量（在失败时抛出通知）。因此，而不是 $ A [指数] 不要 $一个['指数'] 。另请参阅为什么$ foo的[bar]错了？

  • Remember to always use quotes when you are trying to access an element of an associative array. Without quotes, PHP tries to interpret it as a constant (throwing a Notice on failure). So instead of $a[index] do $a['index']. Please also see Why is $foo[bar] wrong?

    在您的code， $ P ['PET'] 将宠物名称的数组，而不是一个爱称。与 in_array测试（）不会成功的，因为它会试图找到在 $过滤器整个阵列。在我的code示例，我使用 和array_diff（） ，以找到两个阵列之间的差异，然后我把它比原来的数组。如果它们匹配，则 $过滤器宠物都没有发现。

    In your code, $p['PET'] will be an array of pet names, not one pet name. Testing with in_array() won't be successful, because it will try to find the whole array in $filter. In my code example, I used array_diff() which will find the difference between the two arrays, then I compare it to the original array. If they match, the $filter pets were not found.

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