duanjiashai9140 2017-12-07 11:51 采纳率: 0%
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Foreach语法问题已经使用了不同项目中的代码现已破解[关闭]

enter image description here

So my problem is I do not understand why I am getting this error as I have used this code previously for a different task and it worked. Here is the sample code:

  <?php

  include('assets/php/sql/con.php');

  if (isset($_GET['age'])) {
    $AGE = $_GET['age'];
    $GENDER = $_GET['gender'];

      $results = DB::query('SELECT * FROM gift_db WHERE age =:age',array(':age' => $_GET['age']))[0]['age'];

      // DB::query('SELECT * FROM gift_db WHERE age=:age', array(':age'=>$_GET['age']));
      $allResults = "";
      foreach ($results as $r) {

        $allResults = $r;

    }
  }
   ?>

If anyone would be able to help out that would be amazing. Thanks :)

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2条回答 默认 最新

  • dongque6377 2017-12-07 12:01
    关注

    When use query function this function returned an array but you selected the first element of this array.

    $results = DB::query('SELECT * FROM gift_db WHERE age =:age',array(':age' => $_GET['age']))[0]['age'];

    Can you try with:

    $results = DB::query('SELECT * FROM gift_db WHERE age =:age',array(':age' => (int) $AGE));

    It's possible in your code insert any value on age.

    PD:Sorry for my English

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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