在html select选项中显示来自sql的值不显示值 [英] displaying values from sql in html select option not displaying values

问题描述

我已经使用php和sql创建了一个html表单.该表单包含一个称为空格的字段，该字段是一个选择下拉列表.这些值来自我在php中声明的数组.如果数组中的值已经存在于数据库中，则不应显示.所以我完成了以下代码:

i have created a html form using php and sql. The form contains a field called spaces which is a select dropdown. The values are coming from an array which i declared in php. If the values in array are already present in database it should not display. So i have done the following code:

<select class="form-control" id="space" name="space">
  <option value="--Select--">--Select--</option>
  <?php
$select=mysqli_query($con,"select * from clients where Spaces IS NOT NULL");
while($menu1=mysqli_fetch_array($select))
      {
$filled =$menu1['name'];
$valuez = array("C101","C102","C103","C104");
if ($filled != $valuez) {


      ?>
    <option value="<?php echo $valuez;?>">
      <?php echo $valuez;?>
    </option>
    <?php
      }}
      ?>
</select>

但这不会显示任何值.谁能告诉我代码中有什么问题吗?谢谢你提前

but this is not making any values display. Can anyone please tell me what is wrong in my code. thanks n advance

推荐答案

您正在将字符串与数组进行比较.您应该像这样使用in_array

you are comparing a string with the array. you should use in_array like this

 <select class="form-control" id="space" name="space">
   <option value="--Select--">--Select--</option>
 <?php
  $select=mysqli_query($con,"select * from clients where Space IS NOT NULL");
 while($menu1=mysqli_fetch_array($select))
  {
  $filled =$menu1['Space'];
 $valuez = array("C101","C102","C103","C104");
 foreach($valuez as $value){
    if($value != $filled){ 
    ?>
        <option value="<?php echo $value;?>">
          <?php echo $value;?>
        </option>
    <?php 
    }
 }
}
?>

更新代码

这篇关于在html select选项中显示来自sql的值不显示值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！