Question

我有一个包含多个字段的表单：Textinputs，Checkboxes，Radios ..我想将它提交到MySQL数据库。当我评论复选框HTML和相应的PHP代码时，一切正常，一切都提交并保存在数据库中。如果我尝试提交复选框表单并取消注释，则不会提交任何内容，单击提交按钮也不会产生任何影响。

如何将checkbox-field的值作为字符串提交给MySQL-Database，并用分号分隔值？对于前者如果复选框字段是：Ab，Cd，De，Fg - 和＆＃34; Ab＆＃34;和＆＃34; De＆＃34;检查后，提交以下字符串：＆＃34; Ab; Cd＆＃34;

以下是我的HTML表单的一部分：

<div class="row">
    <div class="col-sm-4">
        <fieldset class="form-group">
            <label for="plattform">Platform</label>
         <form id="formId">


                <input type="checkbox" name="check_list[]" value="Android">Android
                <input type="checkbox" name="check_list[]" value="iPhone">iPhone
                <input type="checkbox" name="check_list[]" value="iPad">iPad
                <input type="checkbox" name="check_list[]" value="Windows Phone">Windows Phone

            </form> 

            <!-- <input type="checkbox" name="check_list[]" value="Android">
            <input type="checkbox" name="check_list[]" value="iPhone">
            <input type="checkbox" name="check_list[]" value="iPad">
            <input type="checkbox" name="check_list[]" value="Windows Phone"> --> 

        </fieldset>
    </div>
    <div class="col-sm-4">
        <fieldset class="form-group">
            <label for="featured">Featured</label>
                <div>
                        <input type="radio" name="featured" required>True</input>
                    </div>
                    <div>
                    <input type="radio" name="featured" required checked>False</input> 
                </div>
        </fieldset>
    </div>

这是我的php文件示例：

<?php /* Attempt MySQL server connection. Assuming you are running
MySQL server with default setting (user 'root' with no password) */

/* Database connection start */ 
$servername = "localhost"; 
$username = "serverName_Here"; 
$password = "password_Here"; 
$dbname = "dbName_Here";

$conn = mysqli_connect($servername, $username, $password, $dbname) or die("Connection failed: " . mysqli_connect_error());

// Check connection 
if($conn === false){
    die("ERROR: Could not connect. " . mysqli_connect_error()); 
}   
// Escape user inputs for security 
$stName = mysqli_real_escape_string($conn, $_POST['sName']); 
$lgName = mysqli_real_escape_string($conn, $_POST['lName']);
$desc = mysqli_real_escape_string($conn, $_POST['desc']);

$Platform = ''; 
if(!empty($_POST['check_list'])) {  
    $counter = 0;
    foreach($_POST['check_list'] as $check) {
        if ($counter < 1) {             
            $Platform = $check;         
        } else {                
            $Platform = $excludePlatform + ';' + $check;        
        }       
        counter++;
    } 
} 
$Platform = mysqli_real_escape_string($conn, $_POST['check_list']);


$sql = "INSERT INTO tableName_Here (stName, lgName, details_description, Platform) VALUES ('$stName', '$lgName', '$desc', '$Platform')"; 

if(mysqli_query($conn, $sql)){
    echo "Records added successfully."; 
} else {
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($conn); 
}   // close connection 
mysqli_close($conn); ?>

Answer 1

您需要在表单标记中设置复选框。

最简单的方法是将表单打开和关闭选项卡放在其余字段代码的上方和下方。提交按钮也应该在表单中。

在后端复选框值将作为数组出现。如果你想用逗号分隔它们，你可以做这样的事情。

$values = implode(', ', $_POST['check_list']);

Answer 2

首先，为什么您还没有使用POST作为form标记中的方法，您还没有使用提交按钮。这样您就可以访问复选框的值了php喜欢

$value=$_POST['check_list[]'];

Answer 3

试试这整个例子，

表格结构

CREATE TABLE IF NOT EXISTS `games` (
  `id` int(12) NOT NULL AUTO_INCREMENT,
  `game_name` varchar(255) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=1;


<?php
include_once("yourconfig.php"); //include your db config file
extract($_POST);
$check_exist_qry="select * from games";
$run_qry=mysql_query($check_exist_qry);
$total_found=mysql_num_rows($run_qry);
if($total_found >0)
{
    $my_value=mysql_fetch_assoc($run_qry);
    $my_stored_game=explode(',',$my_value['game_name']);
}

if(isset($submit))
{
    $all_game_value = implode(",",$_POST['games']);
    if($total_found >0)
    {
        //update
        $upd_qry="UPDATE games SET game_name='".$all_game_value."'";
        mysql_query($upd_qry);

    }
    else
    {
        //insert
        $ins_qry="INSERT INTO games(game_name) VALUES('".$all_game_value."')";
        mysql_query($ins_qry);
    }
}

?>
<form method="post" action="">
Games You Like: <br/>
    <input type="checkbox" name="games[]" value="1" <?php if(in_array(1,$my_stored_game)){echo "checked";}?>><label>Football</label><br>
    <input type="checkbox" name="games[]" value="2" <?php if(in_array(2,$my_stored_game)){echo "checked";}?>><label>Basket Ball</label><br>
    <input type="checkbox" name="games[]" value="3" <?php if(in_array(3,$my_stored_game)){echo "checked";}?>><label>Pool</label><br>
    <input type="checkbox" name="games[]" value="4" <?php if(in_array(4,$my_stored_game)){echo "checked";}?>><label>Rugby</label><br>
    <input type="checkbox" name="games[]" value="5" <?php if(in_array(5,$my_stored_game)){echo "checked";}?>><label>Tennis</label><br>
    <input type="checkbox" name="games[]" value="6" <?php if(in_array(6,$my_stored_game)){echo "checked";}?>><label>Cricket</label><br>
    <input type="checkbox" name="games[]" value="7" <?php if(in_array(7,$my_stored_game)){echo "checked";}?>><label>Table Tennis</label><br>
    <input type="checkbox" name="games[]" value="8" <?php if(in_array(8,$my_stored_game)){echo "checked";}?>><label>Hockey</label><br>
    <input type="submit" name="submit" value="submit">
</form>

这只是我在本例中添加的基本示例和查询，您可以从这个基本示例中学习，我认为这对您非常有用...如果对此解决方案给出正确答案有用

使用php

3 个答案: