在foreach循环中使用PHP解析数组值

Question

我想用PHP的foreach循环解析一个数组，以获取'ques'数组中的对象名称和值。我想要   

[
 {
 "ques": [
{
    "name": "comment",
    "value": "comment me for the reason",
    "sur_id": "1",
    "user_id": "admin@gmail.com",
    "pagename": "question_response"
},
{
    "name": "check-box[]",
    "value": "1"
},
{
    "name": "radio",
    "value": "radio 2"
},
{
    "name": "yes",
    "value": "no"
}

]
"ques":[
{
    "name": "date",
    "value": "2015-10-23"
    "user_id": "admin1@gmail.com",  
},
{
    "name": "select-deopdown",
    "value": ""
},
{
    "name": "true",
    "value": "false"
},
{
    "name": "number",
    "value": "55"
 }
 ]
}
]

我想将名称，值和user_id与'ques'数组分开：

while ($fetch = mysql_fetch_array($query1)) {
$content = $fetch['CONTENT_VALUES'];
// print_r($content);
$content_value= mb_convert_encoding($content ,"UTF-8");
$datas = json_decode($content, true);
 foreach($datas->ques as $values)
 {
     echo $values->value . "\n";
      print_r($values);
 }
 $test[] = array('ques' =>  $datas ,'answer'=>$values);
}

Answer 1

如果要为每个值创建三个不同的数组，则创建三个空白数组，然后在foreach中将相应的值存储到它们中。我给你一个常见的例子

$name = array(); $values= array(); $users = array();   foreach($datas->ques as $values) {
 $name[] = $values->name;   $values[] = $values->value; $users[] = !empty($values->user_id) ? $values->user_id : '';
  }  

并且您可以根据需要进行修改。

谢谢。

Answer 2

var my_layer = boxLayer;
if (my_layer instanceof OpenLayers.Layer.Vector && my_layer.features.length == 0)
      {
        console.log('Object not ready :( ')
        return;
      }
    console.log('projection (srsName) =', my_layer.features[0].layer.projection.getCode());//"EPSG:3857"
    layer['format'] = new OpenLayers.Format.GML();
    console.log('OpenLayers.Format.GML:\n ',my_layer.format.write(my_layer.features[0]));//GML string without srsName

Answer 3

This could help

$data = json_decode($json); $result = []; foreach($data as $row) {  foreach($row as $k => $v) { $i = 0; foreach($v as $key => $val) { foreach($val as $k1 => $v1) { $result[$i][$k1] = $v1; }
$i++; } } }

echo json_encode($result);

Answer 4

我通过以下代码解决了这个问题。感谢他们中的3个人通过提供想法来指导。

$datas = json_decode($content, true);
foreach($datas as $values)
 {
 $name = $values['name'];
 $value = $values['value'];
 $user_id = $values['user_id'];
 $test[]= array('user'=>$user,'name'=>$name,'value'=>$value);
 } 
}