如何将json对象内嵌到php中的字符串? [英] How do I implode a json object to a string in php?

问题描述

我将这个json保留在要转换为字符串的变量中吗?

I have this json held in a variable that I am trying to convert to a string?

$json_object = '[{"125":"1"},{"126":"2"},{"127":"3"},{"128":"4"},{"129":"5"},{"130":"6"},{"131":"7"},{"132":"8"},{"133":"9"},{"134":"10"},{"135":"11"}]';

我想要的最终结果是这样的东西:

What I want as my end result is something that looks like this:

$json_object1 = '1,2,3,4,5,6,7,8,9,10,11';
$json_object2 = '125,126,127,128,129,130,131,132,133,134,135';

有没有一种方法可以修改implode(",",$json_object)函数来实现这一目标?

Is there a way we can modify the implode(",",$json_object) function to achieve this?

另一个问题: 任何想法我们怎么可能把它

Another question: Any idea how we might put this

{"26":"Child - 1500.00","28":"Foreigner - 4000.00","27":"Resident - 3000.00"}

到类似列表

26 : Child - 1500.00
27: Resident - 3000.00
28: Foreigner - 4000.00

推荐答案

$json_object = '[{"125":"1"},{"126":"2"},{"127":"3"},{"128":"4"},{"129":"5"},{"130":"6"},{"131":"7"},{"132":"8"},{"133":"9"},{"134":"10"},{"135":"11"}]';

$json = json_decode($json_object);
echo implode(", ", array_map(function($obj) { foreach ($obj as $p => $v) { return $p;} }, $json));
echo "<br>";
echo implode(", ", array_map(function($obj) { foreach ($obj as $p => $v) { return $v;} }, $json));

请参见 https://3v4l.org/p3p45

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