Question

我有一个数据库，我想插入用户选择的饮料。如果他们选择饮用其中一种饮料，我希望将1号分配给记录。

HTML

<form role="form" action="/form.php" name="submitfkgo" id="submitfkgo" method="post" onsubmit="return submitfk()">
<input type="email" class="form-control" id="FormEmail" name="FormEmail" placeholder="Enter email">
<input type="text" class="form-control" id="FormName" name="FormName" placeholder="Enter full name">
<input type="checkbox" name="checker[]" id="formOpt1" value="A">
Yes, I will drink A
<input type="checkbox" name="checker[]" id="formOpt2" value="B">
Yes, I will drink B
<input type="checkbox" name="checker[]" id="formOpt3" value="C">
Yes, I will drink C
<a href="#join" onclick="javascript:submitfk();return false" class="btn btn-primary btn-lg btn-block">Submit my application!</a>
</form>

PHP

以下PHP连接到数据库。然后它将值安全地存储到变量中。然后它创建第一条记录。然后我想用他们从上面的选择中勾选的内容更新记录。目前，所有组合都只用0更新。

<?php
// DB
include("../inc/c.php");
mysql_select_db("join", $con);

// VARS
$FormEmail=mysql_real_escape_string($_POST['FormEmail']);
$FormName=mysql_real_escape_string($_POST['FormName']);
$chkbox = array('A', 'B', 'C');
$choiceDrink=mysql_real_escape_string($_POST['checker']);

// SQL
$r=mysql_query("
    INSERT INTO `db`.`join` (`id`, `FormEmail`, `FormName`) VALUES (NULL, '".$FormEmail."', '".$FormName."');
");

// GET INSERTED RECORD
$id = mysql_insert_id();

$values = array();
foreach($chkbox as $selection ){
    if(in_array($selection, $choiceDrink))
    {
        $values[ $selection ] = 1;
    } else {
        $values[ $selection ] = 0;
    }
}

// SQL FOR CHOICES
$r2=mysql_query("
    UPDATE `db`.`join`
        SET
        `A` = '".$values['A']."',
        `B` = '".$values['B']."',
        `C` = '".$values['C']."'
    WHERE
        `join`.`id` =".$id.";
");

// FORM
echo '<script type="text/javascript">window.location="/thanks"</script>';

// MYSQL CLOSE
mysql_close($con);
?>

Answer 1

请勿在{{1}}上使用mysql_real_escape_string。当您将输入直接替换为数据库时，您只需要这样做。所以它应该是：

$_POST['checker']

Answer 2

我建议，主要是检查$_POST['checker']的数组大小。然后有条件地将SQL查询构建为

if(count($_POST['checker']) ==0)
{
// SQL Query

}
else if (count($_POST['checker']) ==1)
{
// SQL query
}

//...
else
{

}

Answer 3

试试这个：

<强> HTML

<form role="form" action="/form.php" name="submitfkgo" id="submitfkgo" method="post" onsubmit="return submitfk()">
<input type="email" class="form-control" id="FormEmail" name="FormEmail" placeholder="Enter email">
<input type="text" class="form-control" id="FormName" name="FormName" placeholder="Enter full name">
<input type="checkbox" name="checker" id="formOpt1" value="A">
Yes, I will drink A
<input type="checkbox" name="checker" id="formOpt2" value="B">
Yes, I will drink B
<input type="checkbox" name="checker" id="formOpt3" value="C">
Yes, I will drink C
<a href="#join" onclick="javascript:submitfk();return false" class="btn btn-primary btn-lg btn-block">Submit my application!</a>
</form>

<强> PHP

...

// VARS
$FormEmail=mysql_real_escape_string($_POST['FormEmail']);
$FormName=mysql_real_escape_string($_POST['FormName']);
$choiceDrink=$_POST['checker'];

// SQL
$r=mysql_query("
    INSERT INTO `db`.`join` (`id`, `FormEmail`, `FormName`) VALUES (NULL, '".$FormEmail."', '".$FormName."');
");

// GET INSERTED RECORD
$id = mysql_insert_id();

$values = array('A' => 0, 'B' => 0, 'C' => 0);
if (isset($values[$choiceDrink])) $values[ $choiceDrink ] = 1;

// SQL FOR CHOICES
$r2=mysql_query("
    UPDATE `db`.`join`
        SET
        `A` = '".$values['A']."',
        `B` = '".$values['B']."',
        `C` = '".$values['C']."'
    WHERE
        `join`.`id` =".$id.";
");

...

尝试使用HTML表单将多个复选框值插入PHP

3 个答案: