JS/PHP - 根据以前的选择自动填充选择字段

因此，我有3个选择字段，我想根据＃1中选定的值填充＃2和＃3。JS/PHP - 根据以前的选择自动填充选择字段

JS正在接受＃1中的字段变更，将其成功提交给我的PHP脚本，但在我的查询中插入$ _POST ['problem']变量时，出现“数组到字符串转换”错误得到相关结果。

我曾尝试json_encode，并在第一选择字段（选择一个id =“问题”）是一个数组，确保原选定的值，因此不明白的地方转换误差的来源。

错误：

<b>Notice</b>: Array to string conversion in <b>Z:\xampp\htdocs\qcisource\ihg\ajax.php</b> on line <b>17</b><br /> 
{"current_field":null,"field_count":null,"lengths":null,"num_rows":null,"type":null}

有什么建议？

HTML

 <p>Problem Experienced</p> 
     <!-- Problem --> 
     <div class="input-group"> 
      <span class="input-group-addon"><i class="fa fa-warning"></i></span> 
      <select id="problem" name="problem" class="form-control select2" multiple="multiple" data-placeholder="Select a Problem"> 
       <option value=""> </option> 
       <?php 

       // define query 
       $sql = "SELECT Issue, Description FROM qci_problems_index_new ORDER BY Issue"; 

       // query 
       $result = $mysqli->query($sql) or die('<p>Query to get Issue from qci_problems_index_new table failed: ' . mysql_error() . '</p>'); 

       while ($row = $result->fetch_array(MYSQLI_ASSOC)) { 
        $problem = $row['Issue']; 
        $desc = $row['Description']; 
        echo "<option value=\"$problem\" data-desc=\"$desc\">" . $problem . "</option>\n"; 
       } 

       $result->free(); 
       ?> 

      </select> 
     </div> 

     <p>Problem Category</p> 
     <!-- Problem Category --> 
     <div class="input-group"> 
      <span class="input-group-addon"><i class="fa fa-warning"></i></span> 
      <select id="problem_category" name="problem_category" class="form-control select2" multiple="multiple" data-placeholder="Select a Problem Category"> 
       <option value=""> </option> 
       <?php 

       // define basic query 
       $sql = "SELECT DISTINCT Category FROM qci_problems_index_new ORDER BY Category"; 

       // query 
       $result = $mysqli->query($sql) or die('<p>Query to get Category data from qci_problems_index_new table failed: ' . mysql_error() . '</p>'); 

       while ($row = $result->fetch_array(MYSQLI_ASSOC)) { 
        $category = $row["Category"]; 
        echo "<option value=\"$category\">" . $category . "</option>\n"; 
       } 

       $result->free(); 
       ?> 
      </select> 
     </div> 

     <p>Department Responsible</p> 
     <!-- Department Responsible --> 
     <div class="input-group"> 
      <span class="input-group-addon"><i class="fa fa-bars"></i></span> 
      <select id="department" name="department" class="form-control select2" multiple="multiple" data-placeholder="Select a Department"> 
       <option value=""> </option> 
       <?php 

       // define basic query 
       $sql = "SELECT DISTINCT Department_Responsible FROM qci_problems_index_new ORDER BY Department_Responsible"; 

       // query 
       $result = $mysqli->query($sql) or die('<p>Query to get department_responsible from qci_problems_index_new table failed: ' . mysql_error() . '</p>'); 

       while ($row = $result->fetch_array(MYSQLI_ASSOC)) { 
        $dept = $row["Department_Responsible"]; 
        echo "<option value=\"$dept\">" . $dept . "</option>\n"; 
       } 

       $result->free(); 
       ?> 
      </select> 
     </div>

<script type="text/javascript" language="javascript"> 
$(function() { 
    $('#problem').change(function() { 
     $.ajax({ 
      type: 'POST', 
      url: 'ajax.php', 
      data: { 
       problem: $(this).val() 
      }, 
      dataType: 'json', 
      success: function (data) 
      { 
       var Category = data[0]; 
       var Department_Responsible = data[1]; 
       $('#problem_category').val(Category); 
       $('#department').val(Department_Responsible); 
      } 
     }); 
    }); 
}); 
</script>

ajax.php

<?php 

if(isset($_POST['problem'])) { 

    // Start MySQLi connection 
    include './plugins/MySQL/connect_db.php'; 
    $db = new mysqli($dbhost,$dbuser,$dbpass,$dbname); 

    // display error if connection cannot be established 
    if($db->connect_errno > 0){ 
    die('Unable to connect to database [' . $mysqli->connect_error . ']'); } 

    // sanitize variables 
    $problem = $_POST['problem']; //mysqli_real_escape throws error, ignore for now 

    // run query 
    $result = $db->query("SELECT Category, Department_Responsible FROM qci_problems_index_new WHERE Issue= '".$problem."'"); 

    // return data as array 
    $array = mysqli_fetch_array($result); 
    echo json_encode($result); 

} 
?>

来源

2016-10-23 Armitage2k

请添加错误的详细信息... –

哇，真的没人......？我在这里偶然发现了一个谜题，或者代码没问题，我的服务器在这里有问题吗？ – Armitage2k

固定它自己...
除了与JavaScript的一些小错误，主要问题是我正在使用不处理AJAX的Jquery Select2插件与常规的<select>字段相同。我必须在我的所有字段上手动trigger('change')才能显示AJAX值。

的Javascript

<script type="text/javascript" language="javascript"> 
$(function() { 
    $('#problem').change(function() { 
     $.ajax({ 
      type: 'POST', 
      url: 'ajax.php', 
      data: 'problem=' + $(this).val(), 
      dataType: 'json', 
      success: function (data) 
      { 
      var data0 = data[0]; 
      var data1 = data[1]; 
      $('#problem_category').val(data0); 
      $('#department').val(data1); 

      $('#problem_category').trigger('change'); 
      $('#department').trigger('change'); 
      } 
     }); 
    }); 

}); 
</script>

Ajax.php

<?php 
error_reporting(E_ALL); 
ini_set('display_errors', 1); 

if(isset($_POST['problem'])) { 

    // Start MySQLi connection 
    include '../../plugins/MySQL/connect_db.php'; 
    $db = new mysqli($dbhost,$dbuser,$dbpass,$dbname); 

    // display error if connection cannot be established 
    if($db->connect_errno > 0){ 
    die('Unable to connect to database [' . $db->connect_error . ']'); } 

    // run query 
    $sql = "SELECT Category, Department_Responsible FROM qci_problems_index_new WHERE Issue= '".$_POST['problem']."'"; 
    $result = $db->query($sql) or die(mysqli_error()); 

    // return data as array 
    $array = mysqli_fetch_array($result); 
    echo json_encode($array); 
} 
?>

HTML：

<p>Problem Experienced</p> 
<!-- Problem --> 
<div class="input-group"> 
    <span class="input-group-addon"><i class="fa fa-warning"></i></span> 
    <select id="problem" name="problem" class="form-control select2" data-placeholder="Select a Problem" style="width: 100%;"> 
     <option value=""> </option> 
     <?php 

     // define query 
     $sql = "SELECT Issue, Description FROM qci_problems_index_new ORDER BY Issue"; 

     // query 
     $result = $mysqli->query($sql) or die('<p>Query to get Issue from qci_problems_index_new table failed: ' . mysql_error() . '</p>'); 

     while ($row = $result->fetch_array(MYSQLI_ASSOC)) { 
      $problem = $row['Issue']; 
      $desc = $row['Description']; 
      echo "<option value=\"$problem\" data-desc=\"$desc\">" . $problem . "</option>\n"; 
     } 

     $result->free(); 
     ?> 

    </select> 
</div> 
<p class="help-block" width="100%"><div id="output01" name="output01"></div></p> 

<hr /> 

<p>Problem Category</p> 
<!-- Problem Category --> 
<select id="problem_category" name="problem_category" class="form-control select2" data-placeholder="Select a Problem Category" style="width: 100%;"> 
    <option value=""> </option> 
    <?php 

    // define basic query 
    $sql = "SELECT DISTINCT Category FROM qci_problems_index_new ORDER BY Category"; 

    // query 
    $result = $mysqli->query($sql) or die('<p>Query to get Category data from qci_problems_index_new table failed: ' . mysql_error() . '</p>'); 

    while ($row = $result->fetch_array(MYSQLI_ASSOC)) { 
     $category = $row["Category"]; 
     echo "<option value=\"$category\">" . $category . "</option>\n"; 
    } 

    $result->free(); 
    ?> 
</select> 

<hr /> 

<p>Department Responsible</p> 
<!-- Department Responsible --> 
<div class="input-group"> 
    <span class="input-group-addon"><i class="fa fa-bars"></i></span> 
    <select id="department" name="department" class="form-control select2" data-placeholder="Select a Department" style="width: 100%;"> 
     <option value=""> </option> 
     <?php 

     // define basic query 
     $sql = "SELECT DISTINCT Department_Responsible FROM qci_problems_index_new ORDER BY Department_Responsible"; 

     // query 
     $result = $mysqli->query($sql) or die('<p>Query to get department_responsible from qci_problems_index_new table failed: ' . mysql_error() . '</p>'); 

     while ($row = $result->fetch_array(MYSQLI_ASSOC)) { 
      $dept = $row["Department_Responsible"]; 
      echo "<option value=\"$dept\">" . $dept . "</option>\n"; 
     } 

     $result->free(); 
     ?> 
    </select> 
</div>

来源

2016-10-27 09:21:59 Armitage2k

JS/PHP - 根据以前的选择自动填充选择字段

回答

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