我是php的新手,我试过这个编码我在下拉列表中选择一个值我想要更新相应的值,我有我的数据库中的用户名列表和他们的ID,我正在显示用户名和我想更新的时候我写了一个sql查询来查找成员id并更新到数据库但是它插入了一个空值。这是我的代码。
下拉列表代码
<?
session_start();
if(!isset($_SESSION[''])){
header("location:");
}
?>
<?php include('dbconnect.php'); ?>
<?php
$ed=$_GET['ed'];
$query=mysql_query("select * from table1 where id='$ed'");
$query2= "select * from table2";
$row=mysql_fetch_assoc($query);
if($_POST['Submit'])
{
$mem= $_POST['memid'];
$memname =mysql_query("select memid from table2 where name='$mem'");
$memname1= mysql_fetch_assoc($memname);
$tot_count = mysql_fetch_assoc($ro_count);
$date=date("d-m-Y");
$status="Active";
$onamo=mysql_real_escape_string($_POST['onamo']);
$heid = mysql_real_escape_string($_POST['memname1']);
if($_POST['heid']=='')
{
$namo1="*Required";
$ok=1;
}
if($_POST['onamo']=='')
{
$onamo1="*Required";
$ok=1;
}
$insert=mysql_query("update table1 set oname='$onamo', heid='$heid' where id='$ed'") or die('error');
if($insert)
{
header("Location");
}
}
?>
<body>
<div id="main_container"><br />
<div class="main_content">
<div class="center_content">
<div class="right_content">
<div class="form">
<form action="" method="post" name="fomo" enctype="multipart/form-data" onsubmit="return fall();" class="niceform">
<h1 align="center">Edit Referal Partner </h1>
<?
if($_GET['val']==1) { echo "<h1 class='FeatureBlockHeader' >Member Added Successfully</h1>"; } ?>
<fieldset>
<dl><dt><label for="Owner Name">Referal Partner Name</label></dt><dd><input name="onamo" type="text" size="53" id="onamo" value="<?=$row['oname']?>"/><b style="color:#CA0000"><?=$onamo1?></b></dd></dl>
<dl><dt><label for="">Health Executives</label>
<?php $result1 = mysql_query($query2);
echo'<select name="memid" >';
while($row = mysql_fetch_assoc( $result1 )) {
echo '<option value="'.$row['name'].'">' . $row['name'] . '</option>';
}
echo '</select>'; ?>
</b></dd></dt>
<dl><dt><label for="submit"></label></dt><dd> <input type="submit" name="Submit" value="Submit"></dd></dl></fieldset>
</table>
</form>
'
我的数据库用空字符串更新,如果我直接传递下拉值名称它正在更新。但我想将相应的memberid更新到我的表。请帮帮我。
答案 0 :(得分:0)
第1阶段:
如果字段为空,则不执行任何操作。 (另外,你的逻辑错误与$ ok)。
建议的代码是:
$ok = 1; // assume ok unless we have an error
if($_POST['heid']=='')
{
$namo1="*Required";
$ok=0; // Set to "0" to say "Not Ok"
}
if($_POST['onamo']=='')
{
$onamo1="*Required";
$ok=0; // Set to "0" to say "Not Ok"
}
if ($ok)
{
// Do your update
$insert = mysql_query("update table1 set oname='$onamo', heid='$heid' where id='$ed'") or die('error');
if($insert)
{
header('location: ???');
exit(); // ALWAYS exit after a header redirect, otherwise the rest of the code will continue to work, then the redirect happens!
}
$ok = 0;
$error = 'Failed to update database'
}
// If you get here, you have an error condition.
**第2阶段:**
在获取变量之前,您应该检查isset($_POST['onamo'])
。否则它会发出警告。这可能会给你错误。你有'heid'和'memname1'之间的差异! :)
$ok = 1; // assume ok unless we have an error
if(!isset($_POST['heid']) || $_POST['heid']=='') // Or is it $_POST['memname1']?
{
$namo1="*Required";
$ok=0; // Set to "0" to say "Not Ok"
}
if(!isset($_POST['onamo']) || $_POST['onamo']=='')
{
$onamo1="*Required";
$ok=0; // Set to "0" to say "Not Ok"
}
if ($ok)
{
$onamo=mysql_real_escape_string($_POST['onamo']);
$heid = mysql_real_escape_string($_POST['memname1']); // Or is it $_POST['heid'] ??
// Do your update
$insert = mysql_query("update table1 set oname='$onamo', heid='$heid' where id='$ed'") or die('error');
if($insert)
{
header('location: ???');
exit(); // ALWAYS exit after a header redirect, otherwise the rest of the code will continue to work, then the redirect happens!
}
$ok = 0;
$error = 'Failed to update database'
}
// If you get here, you have an error condition.