Question

我正在对URL执行GET php函数，然后在JSON中返回结果响应：

{
    "results": [{
        "text": "The 2014 BICSI Canadian Conference and Exhibition in Vancouver,    British Columbia, Canada is 61 days away on April 27 - 30.",
        "createdAt": "2014-02-  24T19:54:08.707Z",
        "updatedAt": "2014-02-24T19:54:08.707Z",
        "objectId": "ZZrZ9OgyRG"
    }, {
        "text": "Only 33 more days fro the 2014 BICSI Canadian Conference and Exhibition in Vancouver, Canada!",
        "createdAt": "2014-03-24T13:23:56.240Z",
        "updatedAt": "2014-03-24T13:23:56.240Z",
        "objectId": "ZqxJRiHoJo"
    }]
}

我可以执行php json_decode并在下面的网页上显示以下结果

text | The 2014 BICSI Canadian Conference and Exhibition in Vancouver, British Columbia, Canada is 61 days away on April 27 - 30.

createdAt | 2014-02-24T19:54:08.707Z

updatedAt | 2014-02-24T19:54:08.707Z

objectId | ZZrZ9OgyRG

text | Only 33 more days fro the 2014 BICSI Canadian Conference and Exhibition in Vancouver, Canada!

createdAt | 2014-03-24T13:23:56.240Z

updatedAt | 2014-03-24T13:23:56.240Z

objectId | ZqxJRiHoJo

我用来在网页上显示上述结果的php代码是：

$returned_content = get_data('https://api.parse.com/1/classes/Alerts');

$data = json_decode($returned_content, true);

foreach ($data as $array1 => $arrayn) {
    foreach ($arrayn as $k => $v) {
        foreach ($v as $t => $s) {
            echo"<p> $t | $s ";
        }
    }
}

如果我只是想显示＆＃39;文字＆＃39;仅在网页上的键/值信息，我应该如何修改php。我希望在网页上看到的所有内容都是：

text | The 2014 BICSI Canadian Conference and Exhibition in Vancouver, British Columbia, Canada is 61 days away on April 27 – 30.

text | Only 33 more days fro the 2014 BICSI Canadian Conference and Exhibition in Vancouver, Canada!

Answer 1

$data = json_decode($json);
foreach ($data->results as $item) {
    echo '<br>text | '.$item->text;
}

如果你没有＃39;添加第二个json_decode()参数，你可以将你的JSON与StdClass一起使用。

Answer 2

将其包裹在简单的if条件中。

foreach ($data as $array1 => $arrayn) {
    foreach($arrayn as $k => $v) {
        foreach ($v as $t => $s) {
            if($t == 'text') {
                echo"<p> $t | $s ";
            }
        }
   }
}

Answer 3

喜欢这个

$returned_content = get_data('https://api.parse.com/1/classes/Alerts');

$data = json_decode($returned_content, true);

foreach ($data as $array1 => $arrayn) {
foreach($arrayn as $k => $v){
foreach ($v as $t => $s)
{
if($t == "text")
echo"<p> $t | $s ";
}
   }

}

Answer 4

这一行之后：

$data = json_decode($returned_content, true);

您有一个关联数组，因此您可以将循环更改为：

foreach ($data as $array1 => $arrayn) {
    foreach($arrayn as $k => $v) {
        foreach ($v as $t => $s) {
            if('text' === $t) {
                echo"<p> $t | $s ";
            }
        }
   }
}

Answer 5

简化并检查密钥是否存在，以防万一JSON不包含密钥＆＃34; text＆＃34;它将返回通知。

$data = json_decode($str,true);
foreach($data["results"] as $key=>$val){
    if(array_key_exists("text",$val)){
        echo 'text | '.$val["text"];
        echo '<br />';
    }
}

这里$str是你的json字符串。

PHP foreach在嵌套数组上

5 个答案: