我尝试使用PHP解析JSON文件。但我现在被困住了。

这是我JSON文件的内容:

{
    "John": {
        "status":"Wait"
    },
    "Jennifer": {
        "status":"Active"
    },
    "James": {
        "status":"Active",
        "age":56,
        "count":10,
        "progress":0.0029857,
        "bad":0
    }
}

这是我目前为止所做的尝试:

<?php

$string = file_get_contents("/home/michael/test.json");
$json_a = json_decode($string, true);

echo $json_a['John'][status];
echo $json_a['Jennifer'][status];

但是因为我事先不知道名字(如“John”,“Jennifer”)和所有可用的键和值(如“age”,“count”),我认为我需要创建一些foreach循环。

我希望你能举个例子。


当前回答

更标准的回答:

$jsondata = file_get_contents(PATH_TO_JSON_FILE."/jsonfile.json");

$array = json_decode($jsondata,true);

foreach($array as $k=>$val):
    echo '<b>Name: '.$k.'</b></br>';
    $keys = array_keys($val);
    foreach($keys as $key):
        echo '&nbsp;'.ucfirst($key).' = '.$val[$key].'</br>';
    endforeach;
endforeach;

输出为:

Name: John
 Status = Wait
Name: Jennifer
 Status = Active
Name: James
 Status = Active
 Age = 56
 Count = 10
 Progress = 0.0029857
 Bad = 0

其他回答

$json_a = json_decode($string, TRUE);
$json_o = json_decode($string);



foreach($json_a as $person => $value)
{
    foreach($value as $key => $personal)
    {
        echo $person. " with ".$key . " is ".$personal;
        echo "<br>";
    }

}

要遍历多维数组,可以使用RecursiveArrayIterator

$jsonIterator = new RecursiveIteratorIterator(
    new RecursiveArrayIterator(json_decode($json, TRUE)),
    RecursiveIteratorIterator::SELF_FIRST);

foreach ($jsonIterator as $key => $val) {
    if(is_array($val)) {
        echo "$key:\n";
    } else {
        echo "$key => $val\n";
    }
}

输出:

John:
status => Wait
Jennifer:
status => Active
James:
status => Active
age => 56
count => 10
progress => 0.0029857
bad => 0

在代码板上运行

我使用下面的代码转换json到PHP数组, 如果JSON有效,那么json_decode()工作良好,并将返回一个数组, 但如果JSON格式不正确,它将返回NULL,

<?php
function jsonDecode1($json){
    $arr = json_decode($json, true);
    return $arr;
}

// In case of malformed JSON, it will return NULL
var_dump( jsonDecode1($json) );
?>

如果在JSON格式不正确的情况下,你只期望数组,那么你可以使用这个函数,

<?php
function jsonDecode2($json){
    $arr = (array) json_decode($json, true);
    return $arr;
}

// In case of malformed JSON, it will return an empty array()
var_dump( jsonDecode2($json) );
?>

如果JSON格式不正确,你想要停止代码执行,那么你可以使用这个函数,

<?php
function jsonDecode3($json){
    $arr = (array) json_decode($json, true);

    if(empty(json_last_error())){
        return $arr;
    }
    else{
        throw new ErrorException( json_last_error_msg() );
    }
}

// In case of malformed JSON, Fatal error will be generated
var_dump( jsonDecode3($json) );
?>

Try

<?php
$string = file_get_contents("/home/michael/test.json");
$json_a = json_decode($string,true);

foreach ($json_a as $key => $value){
  echo  $key . ':' . $value;
}
?>

试一试:

foreach ($json_a as $key => $value)
 {
   echo $key, ' : ';
   foreach($value as $v)
   {
       echo $v."  ";
   }
}