Question

我有两个复选框

<form action="" method="post" enctype="multipart/form-data">
<input type="checkbox" name="file_type" value="1"> Filer<br />
<input type="checkbox" name="file_type" value="1"> Statistik
</form>

我的数据库表das_custermer_files上有两行。

第1行=文件第2行=统计

我如何检查其中一个将值放入我的行。

我试过这段代码

if($_POST['file_type'] == 1){
            mysql_query("INSERT INTO das_custermer_files (url, das_file_categories_id, das_custermers_id, name) 
                         VALUE('$fileurl', 1, $user_custermers_id, '$name')"
                    ) or die(mysql_error());
            }

            elseif($_POST['statistik_type'] == 1){
                mysql_query("INSERT INTO das_custermer_files (url, das_custermer_upload_id, das_custermers_id, name) 
                         VALUE('$fileurl', 1, $user_custermers_id, '$name')"
                    ) or die(mysql_error());
            }
            else{
                echo "no choices";
            }

但是，如果我只检查其中一个，那么它在我的两行中都保留了值1。

希望人们明白。如果不。问

我做到了。

精细代码：

HTML

<form action="" method="post" enctype="multipart/form-data">
    <input type="radio" name="file_type" value="1"> Filer<br />
    <input type="radio" name="file_type" value="2"> Statistik
</form>

PHP

$field = false;

            switch($_POST['file_type'])
            {
                case 1:
                    $field = 'das_file_categories_id'; 
                break;
                case 2:
                    $field = 'das_custermer_upload_id'; 
                break;
                default:
                $field = false;
            }
            if($field)
            {
               mysql_query("INSERT INTO das_custermer_files (url, $field, das_custermers_id, name) 
                         VALUE('$fileurl', 1, $user_custermers_id, '$name')"
                    ) or die(mysql_error());
            }

Answer 1

我的方法是这样的：

我会改用radio按钮来确保选择一个类型。

表格位：

    <input type="radio" name="file_type" value="0" checked="checked"> None<br />
    <input type="radio" name="file_type" value="1"> Filer<br />
    <input type="radio" name="file_type" value="2"> Statistik

PHP位：

$field = false;

 // set the field name based on the file type selected in the form
switch($_POST['file_type'])
{
    case 1:
        $field = '`das_file_categories_id`';
    break;
    case 2:
        $field = '`das_custermer_upload_id`'; 
    break;
    default:
    $field = false;
}

// if a field has been set ( i.e file_type != 0 ) then build and run the query
if($field)
{

    $query  =   "INSERT INTO `das_custermer_files` (`url`, " . $field . ", `das_custermers_id`, `name`) ";
    $query .=   "VALUES ('" . $fileurl . "', 1, " . $user_custermers_id . ", '" . $name . "')";
    mysql_query($query);
    echo "Inserting to db: " . $query;
} else {
    echo "No field set: ";
    print_r($_POST);
}

值得注意的事情

您使用的mysql_query已过时，请参阅此处：http://php.net/manual/en/function.mysql-query.php
即使您使用$fileurl，$user_custermers_id，$name我也看不到这些设置，所以我假设它们设置在文件中更高的位置

Answer 2

您的check_box名称应该不同。你的一样。

<input type="checkbox" name="file_type" value="1"> Filer<br />
<input type="checkbox" name="file_type" value="1"> Statistik

Answer 3

Html代码：

<form action="test.php" method = 'post'>
<input type="checkbox" name="file_type[]" value="1"> Filer<br />
<input type="checkbox" name="file_type[]" value="2"> Statistik
<input type="submit" value="submit" />

复选框将在数组$ _POST ['file_type']中包含其值你可以按如下方式将其值与db相互匹配： -

if(isset($_POST['file_type'])){
 foreach($_POST['file_type'] as $keys)
   {     
     mysql_query("INSERT INTO das_custermer_files (url, das_file_categories_id,     das_custermers_id, name) 
                 VALUE('$fileurl', $keys, $user_custermers_id, '$name')"
            ) or die(mysql_error());
    }
 }
    else if(isset($_POST['statistik_type'])){
        mysql_query("INSERT INTO das_custermer_files (url, das_custermer_upload_id, das_custermers_id, name) 
                 VALUE('$fileurl', 1, $user_custermers_id, '$name')"
            ) or die(mysql_error());
    }
    else{
        echo "no choices";
    }

这可能是你的要求......

Answer 4

试试这个。应该工作。

 if(isset($_POST['file_type'])){
        mysql_query("INSERT INTO das_custermer_files (url, das_file_categories_id, das_custermers_id, name) 
                     VALUE('$fileurl', 1, $user_custermers_id, '$name')"
                ) or die(mysql_error());
        }

        else if(isset($_POST['statistik_type'])){
            mysql_query("INSERT INTO das_custermer_files (url, das_custermer_upload_id, das_custermers_id, name) 
                     VALUE('$fileurl', 1, $user_custermers_id, '$name')"
                ) or die(mysql_error());
        }
        else{
            echo "no choices";
        }

Answer 5

你接近是完全错误的。你应该遵循这个结构

<form action="test.php" method = 'post'>
    <input type="checkbox" name="file_type" value="1"> Filer<br />
    <input type="checkbox" name="file_type" value="2"> Statistik
    <input type="submit" value="submit" />
</form>

现在用php

$query  =   '';
if($_POST['file_type'] == 1){
    $query  =   "INSERT INTO das_custermer_files 
                (url, das_file_categories_id, das_custermers_id, name) 
                VALUE
                ('$fileurl', 1, $user_custermers_id, '$name')";
}else if($_POST['file_type'] == 2){
    $query  =   "INSERT INTO das_custermer_files 
                (url, das_custermer_upload_id, das_custermers_id, name) 
                VALUE
                ('$fileurl', 1, $user_custermers_id, '$name')";
}

mysqli_query($query);

Answer 6

您可能忘了<form action = "..." method = "POST"> </form>。

第一个步骤，您应该进行测试：

print_r($_POST);

您通过此测试得到了什么？如果第一步是正确（非空数组），我们可以完成：

if (isset($_POST['file_type'])) { 
   ... 
};

// OR use alternative
if ($_POST['file_type'] == 1) { 
   ... 
};

Answer 7

点击一个复选框，不要更改值属性，但在圆顶中添加“已选中”，如下所示：

<input type="checkbox" name="file_type" value="1" checked> Filer<br />

因此，复选框的提交值将永远保持为“1”

修改

使用字符串替换值时，可以在php中检查它。

HTML

<form action="" method="post" enctype="multipart/form-data"> <input type="checkbox" name="file_type" value="Filer"> Filer<br /> <input type="checkbox" name="file_type" value="Statistik"> Statistik </form>

PHP

if($_POST['file_type'] == "Filer"){ mysql_query("INSERT INTO das_custermer_files (url, das_file_categories_id, das_custermers_id, name) VALUE('$fileurl', 1, $user_custermers_id, '$name')" ) or die(mysql_error()); } elseif($_POST['file_type'] == "Statistik"){ mysql_query("INSERT INTO das_custermer_files (url, das_custermer_upload_id, das_custermers_id, name) VALUE('$fileurl', 1, $user_custermers_id, '$name')" ) or die(mysql_error()); } else{ echo "no choices"; }

使用PHP和MYSQL复选框

7 个答案: