如何为 json_decode 的每个元素编写查询？答案

【问题标题】：How can I write a query for each element of json_decode?如何为 json_decode 的每个元素编写查询？
【发布时间】：2016-10-11 21:49:13
【问题描述】：

我在 Android 中将一个 JSON 数组作为字符串发布，并使用 JSON decode 对其进行解码。这是我的 PHP 代码：

 if($_SERVER['REQUEST_METHOD']=='POST'){
     //Getting values
     $courseList = $_POST['courseList'];

     $professor_ID = $_POST['Prof_ID'];

     $list = json_decode($courseList);

     print_r($list);
     require_once('dbConnect.php');

$sql = "select Stud_ID,student.f_Name,student.l_Name from student,course where course.S_ID = student.Stud_ID and course.P_ID in ($list)";
.
.
.
?>

这是 print_r($list) 结果：

 Array
(
[0] => 1
[1] => 2
)

现在，我想为 JSON 解码的每个元素编写一个查询，但它不起作用。

【问题讨论】：

标签： php android mysql json

【解决方案1】：

一个可能的解决方案（如果我理解正确的话）：

implode(', ', $list)

$sql = "select Stud_ID,student.f_Name,student.l_Name from student,course where course.S_ID = student.Stud_ID and course.P_ID in (implode(', ', $list))";

【讨论】：

【解决方案2】：

您可以使用 foreach 循环或内爆

<?php

 if($_SERVER['REQUEST_METHOD']=='POST'){
  //Getting values 
 require_once('dbConnect.php');
 // i am assuming $con is your connection variable

 // its good practice to check against sql injection
 $courseList = mysqli_real_escape_string( $con, $_POST['courseList'] );
 $professor_ID = mysqli_real_escape_string( $con, $_POST['Prof_ID'] );

 $list = json_decode( $courseList );

 print_r( $list );

 // Assuming $list is 
 //   Array
 //  (
 //    [0] => 1
 //    [1] => 2
 //  )    

 foreach ( $list as $p_id ) {
   // prepare your query
   $sql = "select Stud_ID,student.f_Name,student.l_Name from student,course where course.S_ID = student.Stud_ID and course.P_ID = $p_id ";
   // execute and do other stuff
}

// or use implode
$pid = (implode(', ', $list);
$sql = "select Stud_ID,student.f_Name,student.l_Name from student,course where course.S_ID = student.Stud_ID and course.P_ID in $pid ";
// execute and do other stuff


?>

【讨论】：