Question

$json=json_encode($_POST['myData']);
$decoded = json_decode($json,true); 
if (is_array($decoded)) {
    foreach ($decoded as $value) {
        echo $value["name"] . "=" . $value["value"];
    }
}

我为每个字段都有单独的值，我想要在数据库中存储单独的值..

发布是这样的：

myData[0][name]  =  name
myData[0][value] =  Yaseen
myData[1][name]  =  age
myData[1][value] =  22
myData[2][name]  =  gender
myData[2][value] =  male
myData[3][name]  =  skills
myData[3][value] =  Php
myData[4][name]  =  note
myData[4][value] =  sad
Please help...Thanks

var_dump（）结果：

    <br/>
     array(1) { <br/>
      ["myData"]=><br/>
      array(5) {<br/>
      [0]=><br/>
      array(2) {<br/>
      ["name"]=><br/>
      string(4) "name"<br/>
      ["value"]=><br/>
      string(6) "xyz"<br/>
     }<br/>
    [1]=><br/>
     array(2) {<br/>
      ["name"]=><br/>
      string(3) "age"<br/>
      ["value"]=><br/>
      string(2) "22"<br/>
    }<br/>
    [2]=><br/><br/>
    array(2) {<br/>
      ["name"]=><br/>
      string(6) "gender"<br/>
      ["value"]=><br/>
      string(4) "male"<br/>
    }<br/>
    [3]=><br/>
    array(2) {<br/>
      ["name"]=><br/>
      string(6) "skills"<br/><br/>
      ["value"]=><br/>
      string(3) "Php"<br/>
    }<br/>
    [4]=><br/>
    array(2) {<br/>
      ["name"]=><br/>
      string(4) "note"<br/>
      ["value"]=><br/>
      string(6) "asdasd"<br/>
    }<br/>
  }<br/>
}<br/>

Answer 1

试

$json=json_encode($_POST['myData']);
$decoded = json_decode($json,true); 

if (is_array($decoded)) {
  for($i = 0; $i <= count($decoded); $i++) {
        echo $decoded[$i]["name"];

    }
}

php Json_decode获取个人数据的正确方法

1 个答案: