MySQL数据库不会从数据库中更新

输出似乎不错，但是从更新输入不会转嫁到数据库MySQL数据库不会从数据库中更新

edit.php - 从这里所有的值显示正确

<?php 
session_start(); 

$name = $_SESSION['name']; 
$sur = $_SESSION['sur']; 
$pass = $_SESSION['pass']; 

echo $name.' '.$sur.' '.$pass; 

//connect 
$dbh = mysql_connect ("localhost", "xxx_admin", "xxx") 
     or die ('ERROR!'); 
     mysql_select_db ("xxx_database"); 

$query="SELECT * FROM client_info WHERE (first='$name' AND last='$sur' AND password='$pass')"; 
$result=mysql_query($query); 


    $id = mysql_result($result,$i,"id"); 
    $first = mysql_result($result,$i,"first"); 
    $last = mysql_result($result,$i,"last"); 
    $phone = mysql_result($result,$i,"phone"); 
    $mob = mysql_result($result,$i,"mob"); 
    $fax = mysql_result($result,$i,"fax"); 
    $email = mysql_result($result,$i,"email"); 
    $web = mysql_result($result,$i,"web"); 
    $com = mysql_result($result,$i,"com"); 
    $add = mysql_result($result,$i,"add"); 
    $city = mysql_result($result,$i,"city"); 
    $state = mysql_result($result,$i,"state"); 
    $zip = mysql_result($result,$i,"zip"); 
    $zone = mysql_result($result,$i,"zone"); 
    $office = mysql_result($result,$i,"office"); 
    $office_num = mysql_result($result,$i,"office_num"); 
    $ext_mob = mysql_result($result,$i,"ext_mob"); 
    $ext_phone = mysql_result($result,$i,"ext_phone"); 
    $ext_office = mysql_result($result,$i,"ext_office"); 
    $srv = mysql_result($result,$i,"srv"); 
    $stype = mysql_result($result,$i,"stype"); 
    $voip = mysql_result($result,$i,"voip"); 
    $vpass = mysql_result($result,$i,"vpass"); 
    $regDate = mysql_result($result,$i,"regDate"); 
    $acct = mysql_result($result,$i,"acct"); 
    $Nagent = mysql_result($result,$i,"Nagent"); 
    $agents = mysql_result($result,$i,"agents"); 
    $password = mysql_result($result,$i,"password"); 

?> 
<html> 
<head> 
<title></title> 
</head> 

<body> 

<form method="post" action="update.php" name="gen"> 
<b>Personal Info:</b> 
<p>First Name:<input type="text" name="first" size="20" value="<?php echo $first; ?>"/></p> 
<p>Last Name:<input type="text" name="last" size="20" value="<?php echo $last; ?>"/></p> 
<p>Mob:<input type="text" name="mob" size="20" value="<?php echo $mob; ?>"/> 
ext:<input type="text" name="ext_mob" size="4" value="<?php echo $ext_mob; ?>"/></p> 
<p>Phone:<input type="text" name="phone" size="20" value="<?php echo $phone; ?>"/> 
ext:<input type="text" name="ext_phone" size="4" value="<?php echo $ext_phone; ?>"/></p> 
<p>Fax:<input type="text" name="fax" size="20" value="<?php echo $fax; ?>"/></p> 
<p>E-mail:<input type="text" name="email" size="35" value="<?php echo $email; ?>"/></p> 
<p>Address:<input type="text" name="add" size="40" value="<?php echo $add; ?>"/></p> 
<p>City:<input type="text" name="city" size="20" value="<?php echo $city; ?>"/></p> 
<p>State:<input type="text" name="state" size="20" value="<?php echo $state; ?>"/></p> 
<p>Zip Code:<input type="text" name="zip" size="5" value="<?php echo $zip; ?>"/></p> 
<p>Zone:<input type="text" name="zone" size="5" value="<?php echo $zone; ?>"/></p> 
<br> 

<b>Office Info:</b> 
<p>Company:<input type="text" name="com" size="40" value="<?php echo $com; ?>"/></p> 
<p>Office Address:<input type="text" name="office" size="40" value="<?php echo $office; ?>"/></p> 
<p>Office Num #:<input type="text" name="office_num" size="15" value="<?php echo $office_num; ?>"/> 
ext:<input type="text" name="ext_office" size="4" value="<?php echo $ext_office; ?>"/></p> 
<p>Website:<input type="text" name="web" size="30" value="<?php echo $web; ?>"/></p><br> 

Old password: 
<input type="password" size="20" name="oldpassword"> 
New password: 
<input type="password" size="20" name="newpassword"> 
verify new password 
<input type="password" size="20" name="verpassword"> 

<input type="text" size="20" name="id" value="<?php echo $id; ?>"> 
<input type="submit" value="Update Database"> 
</form> 




</body> 
</html>

update.php - 它说更新成功，但也有我的数据库

<?php 
session_start(); 

$ud_id=$_POST['id']; 
$name = $_SESSION['name']; 
$sur = $_SESSION['sur']; 
$pass = $_SESSION['pass']; 

$ud_first = $_POST['first']; 
$ud_last = $_POST['last']; 
$ud_phone = $_POST['phone']; 
$ud_mob = $_POST['mob']; 
$ud_fax = $_POST['fax']; 
$ud_email = $_POST['email']; 
$ud_web = $_POST['web']; 
$ud_com = $_POST['com']; 
$ud_add = $_POST['add']; 
$ud_city = $_POST['city']; 
$ud_state = $_POST['state']; 
$ud_zip = $_POST['zip']; 
$ud_zone = $_POST['zone']; 
$ud_office = $_POST['office']; 
$ud_office_num = $_POST['office_num']; 
$ud_ext_mob = $_POST['ext_mob']; 
$ud_ext_phone = $_POST['ext_phone']; 
$ud_ext_office = $_POST['ext_office']; 
$ud_password = $_POST['newpassword']; 

//connect 
$dbh = mysql_connect ("localhost", "xxx_admin", "xxx") 
     or die ('ERROR!'); 
     mysql_select_db ("xxx_database"); 


$query="UPDATE client_info SET first='$ud_first',last='$ud_last',phone='$ud_phone',mob='$ud_mob',fax='$ud_fax',email='$ud_email',web='$ud_web',com='$ud_com',add='$ud_add',city='$ud_city',state='$ud_state',zip='$ud_zip',zone='$ud_zone',office='$ud_office',office_num='$ud_office_num',ext_mob='$ud_ext_mob',ext_phone='$ud_ext_phone',ext_office='$ud_ext_office',password='$ud_password' WHERE id='$ud_id'"; 

mysql_query($query); 

echo "Record Updated at ID: ".$ud_id; 
mysql_close();  

?>

来源

2012-10-29 telexper

它说成功，因为它没有其他只要数据库连接正常工作。尝试打印出帖子值，并确保你确实获得了你应该在的位置以及严重不安全的代码记录。 –

更改：mysql_query（$ query）; -to- mysql_query（$ query）或者die（mysql_error（））; –

做一件事，回应你的更新查询并将其粘贴到mysql中，看看有什么错误。 – pkachhia

因为我不知道你的表结构，请试试这个并进行相应的调试。

$query="UPDATE client_info SET first='$ud_first', 
last='$ud_last', 
phone='$ud_phone', 
mob='$ud_mob', 
fax='$ud_fax', 
email='$ud_email', 
web='$ud_web', 
com='$ud_com', 
add='$ud_add', 
city='$ud_city', 
state='$ud_state', 
zip='$ud_zip', 
zone='$ud_zone', 
office='$ud_office', 
office_num='$ud_office_num', 
ext_mob='$ud_ext_mob', 
ext_phone='$ud_ext_phone', 
ext_office='$ud_ext_office', 
password='$ud_password' WHERE id=$ud_id;"; 

mysql_query($query) or die(mysql_error());

来源

2012-10-29 06:58:28 Vinay

所有的
没有改变我强烈建议你逃避你用户在向数据库发送任何内容之前输入。它会逃脱特殊的字符和使用关键字来搅乱你的分贝。要这样做，只需使用mysql_real_escape_string（）就可以了;

$ ud_first = mysql_real_escape_string（$ _ POST ['first']）;
你应该检查mysqli，因为mysql被释放。如果你刚刚开始你的代码，我希望你尽快切换。效率和安全性将提高。
加一个;在你查询的结尾。将帮助服务器知道它结束的地方。
更改or die ('ERROR!');通过或die (mysql_error());这将回应最后由mysql服务器encourtered错误。它很可能会告诉你你的问题是什么。

我给你提示，因为我不明白为什么你的代码不能按预期工作。

来源

2012-10-29 05:52:58

三思而后行，它可能只是一个'你的价值观中没有逃脱的东西，搞乱了服务器。 –

主机只提供mysql，在我的hostserver上仍然没有mysqli – telexper

这个问题的范围之外byt为什么要分别分配每个结果值？

你期待一个结果，并为$我没有设置实际上是空它的工作原理也许你可以简化你的生活和代码使用

$r = mysql_fetch_assoc($result); 
foreach($r as $key=>$value){ $$key=$value; }

是否会有预期或可能更多的行，你就必须在foreach循环或类似的解决方案中关闭它。另一边，加工后和保护输入可能很短&简单太

$allowed_post_vars=array('city','zip',...); // define ok variables 
foreach($allowed_post_vars as $postvar)// each of them 
{ $ud='ud_'.$postvar; // prepare longer name, like $ud_city etc 
    $$ud=mysql_real_escape_string($_POST[$postvar]); //assign it sanitized value 
}

我holpe这些技巧可以帮助你简化代码，并期待更好的进入问题

来源

2012-10-29 07:33:57 Peminator

MySQL数据库不会从数据库中更新

回答

相关问题