如何使用php保存数据库中的多个单选按钮而不保存按钮值

问题描述：

how to save multiple radio button in database using php without save button value.

my code :

$user_id = $_POST['user_id'];
foreach ( $_POST as $key => $val ) {   
    if ($key <> 'user_id') {
        $bidder_interst_insert="INSERT INTO bidder_interest_list(id, bidder_id, bidder_interest_name) VALUES ('','$user_id','$val')";
        $bidder_interst_insert_result = mysql_query($bidder_interst_insert);
        if (mysql_affected_rows() > 0) {
            $interest_list_success = "Thank you Successfull insert your interst list.";
            $_SESSION['interest_list_success_msg'] = $interest_list_success;
        } else {
            $insert_error = "interst list Insert Error.";
            $_SESSION['insert_error_msg'] = $insert_error;
            header("location:interest_list.php");
        }
    }
}

This code work but database extra save in save button value how to solved this problem??

如何在没有保存按钮值的情况下使用php保存数据库中的多个单选按钮。 p> 我的代码： p>

  $ user_id = $ _POST ['user_id']; 
foreach（$ _POST as $ key =＆gt; $ val）{
 if（$  key＆lt;＆gt;'user_id'）{
 $ bidder_interst_insert =“INSERT INTO bidder_interest_list（id，bidder_id，bidder_interest_name）VALUES（''，'$ user_id'，'$ val'）”; 
 $ bidder_interst_insert_result = mysql_query（  $ bidder_interst_insert）; 
 if（mysql_affected_rows（）＆gt; 0）{
 $ interest_list_success =“谢谢你成功插入你的第一个列表。”; 
 $ _SESSION ['interest_list_success_msg'] = $ interest_list_success; 
} else {  
 $ insert_error =“interst list Insert Error。”; 
 $ _SESSION ['insert_error_msg'] = $ insert_error; 
 header（“location：interest_list.php”）; 
} 
} 
} 
   code>  pre> 
 
 此代码工作但数据库额外保存在保存按钮值中如何 解决了这个问题??  p> 
 
 
    p> 
  div>

答

foreach ( $_POST as $key => $val ){

You are directly looping the $_POST, so the SAVE button value is also saving in the database. Take the values individually instead of looping the whole $_POST, then the SAVE button value won't be saved in the database.

And moreover you are using mysql functions which are deprecated, use mysqli or PDO.

EDIT:: Just take it the same way as u took user_id ==> $variablename = $_POST['fieldname'];

EDIT::: Let me suppose i have a form like this

<form name="form1" id="form1" method="post" action="">
        <input type="checkbox" name="products[]" value="A" checked="checked" />A <br />
        <input type="checkbox" name="products[]" value="B" checked="checked" />B <br />
        <input type="checkbox" name="products[]" value="C" checked="checked" />C <br />
        <input type="checkbox" name="products[]" value="D" checked="checked" />D <br />
        <input type="checkbox" name="products[]" value="E" checked="checked" />E <br />
        <input type="checkbox" name="products[]" value="F" checked="checked" />F <br />

        <input type="submit" name="save" id="save" value="Save" />
    </form>

then i can do it like:

<?php
    if(isset($_POST['save']))
    {
        $products = $_POST['products'];
        foreach($products as $key => $value)
        {
            $qry = mysql_query("INSERT INTO tbl(product) VALUES('$value')");
        }
    }
?>

答

Try this

unset($_POST['name-of-save-button']); $data = $_POST; foreach ( $data as $key => $val ){//your code here}

如何使用php保存数据库中的多个单选按钮而不保存按钮值

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