Question

我想使用处于选中状态的复选框更新数据库。

使用选择选项进行更新。

仅选中要更新的复选框。

<?php

    $sql = mysqli_query($conn, "SELECT * FROM leads ORDER BY lid ASC");
    if($count = mysqli_num_rows($sql) == 0){
    echo '<tr><td colspan="8">No Data Entry</td></tr>';
    }else{
    while($row = mysqli_fetch_assoc($sql)){
    echo '
    <tr>
        <td> <input name="checkbox[]" value="'.$row['lid'].'" type="checkbox"></td>
        <td>'.$row['name'].'</td>
        <td>'.$row['gender'].'</td>
        <td>'.$row['phone'].'</td>
        <td>'.$row['company'].'</td>
        <td>'.$row['vehicle'].'</td>
    </tr>
    ';
    }
}?>

这是我的选择代码

<select name="name" class="form-control" onchange="form.submit()">
<option value="0">Filter Data Member</option>
<?php

$sql = mysqli_query($conn, "SELECT * FROM list ORDER BY uid ASC");
if($count = mysqli_num_rows($sql) == 0){
    echo '<tr><td colspan="8">No Data Entry</td></tr>';
}else{
    while($row = mysqli_fetch_assoc($sql)){
        echo '<option name="name" value="'.$row ['name'].'">'.$row ['name'].'</option>
        ';
    }
}?>

这是我的更新

<?php 
      if( isset($_POST['submit']) & empty($_POST["checkbox"]) ) { 
        echo'<div class="alert alert-warning alert-dismissable"><button type="button" class="close" data-dismiss="alert" aria-hidden="true">&times;</button> <strong>Oops!!</strong> Please select atlest one.</div>'; }
    elseif(isset($_POST['submit']) & !empty($_POST["checkbox"]) ){
    $name            = $_POST['name'];

    for($i=0;$i<count($_POST['checkbox']);$i++){
        $trans_id=$_POST['checkbox'][$i];
        $sql = "UPDATE leads SET luser='$name' ";
        $result = mysqli_query($conn,$sql) or die(mysqli_error());
    }
    if($result){
    echo "<meta http-equiv=\"refresh\" content=\"0;URL=trans.php\">";
}}?>

使用此代码我的更新执行工作但我的问题不仅仅是复选框更新，而且所有数据库表列表都同时更新。

感谢任何形式的帮助或建议。谢谢......

Php mysqli使用复选框更新数据库

0 个答案: