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在mysqli php中显示同一列的多个值

[英]display multiple value of same column in mysqli php

I want to display value in checkbox.我想在复选框中显示值。 i have multiple value in database as shown in pic above.我在数据库中有多个值,如上图所示。 please have a look.请看看。 I want value in different checkbox not in same checkbox我希望不同复选框中的值不在同一个复选框中

现在看起来如何[![我的桌子。[![预期结果 ] 1 ] 2 ] 1 ] 2

Code I am trying我正在尝试的代码

<div class="list-group">
            <h3>Name</h3>
            <?php 
                $query = "select distinct(name) from info_user where user_status = '1'";  
                $rs = mysqli_query($con,$query) or die("Error : ".mysqli_error());
                while($color_data = mysqli_fetch_assoc($rs)){

            ?>
                <a href="javascript:void(0);" class="list-group-item"> 
                <input type="checkbox" class="item_filter colour" value="<?php echo $color_data['name']; ?>"  >
                &nbsp;&nbsp; <?php echo $color_data['name']; ?></a>
            <?php } ?>  
            </div>

and what i tried by myself以及我自己尝试过的

  <div class="list-group">
            <h3>Name</h3>
            <?php 
                $column = array();
                $query = "select name from info_user where user_status = '1'";  
                $rs = mysqli_query($con,$query) or die("Error : ".mysqli_error());
                while($color_data = mysqli_fetch_assoc($rs)){
                    $column[] = $color_data['name'];


            ?>
                <a href="javascript:void(0);" class="list-group-item"> 
                <input type="checkbox" class="item_filter colour" value="<?php foreach($column as $value)echo $value['name']; ?>"  >
                &nbsp;&nbsp;
                 <?php foreach($column as $value)echo $value['name']; ?></a>
            <?php } ?>  
            </div> 

Getting this error after trying code尝试代码后出现此错误

将新代码放入页面后出错

you can try this way, update your code你可以试试这种方式,更新你的代码

<div class="list-group">
    <h3>Name</h3>
    <?php
    $column = array();
    $query = "select distinct(name) from info_user where user_status = '1'";
    $rs = mysqli_query($con, $query);
    while ($color_data = mysqli_fetch_assoc($rs)) {
        $column = array_merge($column, explode(',',$color_data['name']));
    }
    // to remove repeated names  
    $column = array_filter($column);
    ?>
    <a href="javascript:void(0);" class="list-group-item"> 
        <input type="checkbox" class="item_filter colour" value="<?php
               foreach ($column as $value) {
                   echo $value;
                   ?>">
            &nbsp;&nbsp;
            <?php
            echo $value;
        }
        ?>
    </a>
</div> 

I think this is what you need.我认为这就是你所需要的。

<div class="list-group">
    <h3>Name</h3>
<?php 
$column = array();
$query = "select name from info_user where user_status = '1'";  
$rs = mysqli_query($con,$query);
while ($color_data = mysqli_fetch_assoc($rs)) {
    $column = array_merge($column, explode(',', $color_data['name']));
}
$column = array_unique($column);
foreach ($column as $value) {
?>
        <a href="javascript:void(0);" class="list-group-item"> 
        <input type="checkbox" class="item_filter colour" value="<?php echo $value; ?>" >
        &nbsp;&nbsp;<?php echo $value; ?>
        </a>
<?php } ?>  
</div> 

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