Question

我有一个简单的表单来自mysql db的显示条目，每个条目都有一组单选按钮 - 隐藏，显示和删除。正如我的代码现在所说，当我提交表单时没有任何反应。

这是我的表单代码

<form class='removeform' action='advertremoveedit.php' method='post'>
  <?php
  include '../inc/connect.php';
  $q = mysqli_query($link, "SELECT * FROM adverts ORDER BY id ASC");
  while($row = mysqli_fetch_array( $q )){
    echo"<input type='radio' name='edit[{$row['id']}]' value='remove'>Remove";
    if ($row['status'] ==  1){
      echo"<input type='radio' name='edit[{$row['id']}]' value='hide'>Hide";
    }
  }
    echo"<input type='submit' name='submit' value='Submit' />";
  ?>
</form>

这是处理表单的php

if(isset($_POST['remove'])){
  $chk = (array) $_POST['remove'];
  $p = implode(',',array_keys($chk)); 
  $t = "SELECT * FROM adverts WHERE id IN ($p)";
  $result = $link->query($t);
  $url=mysqli_fetch_array($result);
  $image=$url['file']; 
  if ($result){
    unlink($image);
    $q = "DELETE FROM adverts WHERE id IN ($p)";
    $delete = $link->query($q);
  }
}

//if hide was clicked, change `status` to 1
if(isset($_POST['show'])){
  $edit= $_POST['show'];
  $chk = (array) $_POST['show'];
  $p = implode(',',array_keys($chk)); 
  $t = "SELECT * FROM adverts WHERE id IN ($p)";
  $result = $link->query($t);
  if ($result){
    $q = "UPDATE adverts SET status = 1 WHERE id IN ($p)";
    $show = $link->query($q);
  }
}

编辑...提交代码

 if(isset($_REQUEST['edit'])){
    if($val=="hide"){
      $chk = (array) $_POST['hide'];
      $p = implode(',',array_keys($chk)); 
      $t = "SELECT * FROM adverts WHERE id = ($p)";
      $result = $link->query($t);
      if ($result){
        $q = "UPDATE adverts SET status = 2 WHERE id = ($p)";
        $hide = $link->query($q);}}}

Answer 1

试试这个

<强>形式：

<form class='removeform' action='advertremoveedit.php' method='post'>
  <?php
  include '../inc/connect.php';
  $q = mysqli_query($link, "SELECT * FROM adverts ORDER BY id ASC");
  while($row = mysqli_fetch_array( $q )){
    echo"<input type='radio' name='edit[".$row['id']."]' value='remove'>Remove";
    if ($row['status'] ==  1){
      echo"<input type='radio' name='edit[".$row['id']."]' value='hide'>Hide";
    }
  }
    echo"<input type='submit' name='submit' value='Submit' />";
  ?>
</form>

并提交页面

if(isset($_REQUEST['edit']))
{
    $arr_submit = $_REQUEST['edit'];
    foreach($arr_submit as $id=>$val)
    {
        if($val=="remove")
        {
            // do your remove task for $id
        }
        else if($val=="hide")
        {
            // do you hide task for $id
        }
    }
}

更新：2

if(isset($_REQUEST['edit']))
{
    $arr_submit = $_REQUEST['edit'];
    foreach($arr_submit as $id=>$val)
    {
        if($val=="remove")
        {
            // delete task on $id

              $t = "SELECT * FROM adverts WHERE id='$id' LIMIT 1";
              $result = $link->query($t);
              if($url=mysqli_fetch_array($result))
              {
                    $image=$url['file'];
                    $q = "DELETE FROM adverts WHERE id='$id' LIMIT 1";
                    $delete = $link->query($q);
              }
        }
        else if($val=="hide")
        {
              $q = "UPDATE adverts SET status = 2 WHERE id='$id' LIMIT 1";
              $show = $link->query($q);
        }
        else if($val=="show")
        {
              $q = "UPDATE adverts SET status = 1 WHERE id='$id' LIMIT 1";
              $show = $link->query($q);
        }
    }
}

Answer 2

将数组绑定到IN条件时，这是一个经典问题。

将$p ID绑定到查询中时，会得到以下SQL：

SELECT * FROM adverts WHERE id IN ('1,2,3');

显然，没有广告具有荒谬的身份'1,2,3'！

您有两个选择：

1）为$p中的每个值运行单独的查询。

2）使用直接连接的值动态构建查询，或者作为首选选项，动态数量的占位符并分别绑定每个值。

单击单选按钮，查找是否隐藏，显示或删除并执行操作

2 个答案: