php - 变量未从 PHP 正确插入 mysql

标签 php mysql

这是我的 PHP 代码:

 // Collect data from URL 
$mid = $_GET['m'];

if (isset($_POST['submit'])) 
{ 
$insert = "insert into table SET from_id = '".$loginuser['members_id']."', to_id = '". $mid ."', date = '".$_POST['date']."' ";
    $add_member = mysql_query($insert);     
}

除了 $mid 之外,数据被正确地输入到数据库中

但是如果在我的 HTML 中我把这个:

    <?php print $mid;?>

然后我可以看到 ID 号的打印...所以我知道我的变量 $mid 具有正确的值....我不知道为什么它没有被插入到数据库中。

我也试过这个SQL

$insert = "insert into table SET from_id = '".$loginuser['members_id']."', to_id = "$mid", date = '".$_POST['date']."' ";
    $add_member = mysql_query($insert);

同样的事情......一切正常,除了 $mid 的值没有进入数据库。

我在数据库中的字段设置为 Int(11) 并且列名没有错误..我检查了 5 次...不知道出了什么问题.. thx

完整代码在这里:

<?
ob_start();
include 'datalogin.php';
//checks cookies to make sure they are logged in 
    if(isset($_COOKIE["user"]))
    {
    $username = $_COOKIE["user"]; 
    $pass = $_COOKIE["password"];
    $check = mysql_query("SELECT * FROM members WHERE email = '$username'")or   die(mysql_error()); 
    $loginuser = false;
        while($info = mysql_fetch_array( $check ))
        {   
            if(! $loginuser)
                 { $loginuser = $info; }

            //if the cookie is present but has the wrong password, they are taken to the login page 
            if ($pass != $info['password']) 
            {
                header("Location: login.php"); 
                exit();
            }  
            else //if the cookie is present and doesn'T have the wrong password they are shown the admin area    
            { 
                include 'header.php';
            }
        }
    }

    else //if there is no cookie present
            { 
                header("Location: login.php"); 
                exit();
            }

 // Collects data from images table 
$mid = $_GET['m'];
 $data = mysql_query("SELECT images.image_id, images.members_id, images.image_url, members.members_id, members.name, members.age
FROM members
LEFT JOIN images
ON members.members_id=images.members_id  WHERE members.members_id ='". $mid ."' ") 
 or die(mysql_error()); 
$data2 = mysql_fetch_array( $data );



    if (isset($_POST['submit'])) 
    { 


        $insert = "insert into booking SET from_id = '".$loginuser['members_id']."', to_id = '$mid', date = '".$_POST['date']."'";
        $add_member = mysql_query($insert) or die(mysql_error());       
        header('Location: index.php'); 
        exit();
    }

?>


  <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" >
<table cellspacing='0' id="booking" align="center" width="680">


<tr>
    <td>Date:</td>
    <td><input name="date" type="text"  size="10" maxlength="10"  class="form-field" /> </td>
</tr>

<tr>
    <td>&nbsp;</td>
    <td><input class="submit-button" type="submit" name="submit" value="SEND REQUEST" /></td>
</tr>
</table>
</form>
<br />

这是表结构

CREATE TABLE IF NOT EXISTS `booking` (
  `booking_id` int(11) NOT NULL AUTO_INCREMENT,
  `from_id` int(11) NOT NULL,
  `to_id` int(11) NOT NULL,
  `date` varchar(10) NOT NULL,
  PRIMARY KEY (`booking_id`)
  ) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=17 ;

最佳答案

这样试试

if (isset($_POST['submit'])) 
{ 
$mid = $_GET['m'];
$insert = "insert into table SET from_id = '".$loginuser['members_id']."', to_id = '". $mid ."', date = '".$_POST['date']."' ";
    $add_member = mysql_query($insert);     
}

关于php - 变量未从 PHP 正确插入 mysql,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20088596/

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