php - 使用 PHP 和 MYSQL 上传信息时不需要我的表单字段

标签 php mysql html forms

我需要在上传信息时不需要我的表单字段。这是场景和代码:

场景:用户想要编辑他们的信息,而不更新不同的图像:但显然,当我单击“提交”并将图像字段留空时,它会引发错误,我该如何防止这种情况发生。

我不需要的字段是图像字段(照片)

这里是代码:

<?php
session_start();
$UserName = $_SESSION['UserName'];
require("checkLoginSession.php");
$adminid = $_GET['id'];

//removed connection
$tbl_name="admin"; // Table name 

// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");


echo("Logged In As: $UserName");
echo "<br />";
echo("We are editing Data for ID: $adminid");
echo "<br />";
echo "<a href=test.php>Go back to panel</a>";

$id=$_GET['id'];
// Retrieve data from database 
$sql="SELECT * FROM admin WHERE id='$id'";
$result=mysql_query($sql) or die(mysql_error());

$rows=mysql_fetch_array($result);
?>
<table width="400" border="0" cellspacing="1" cellpadding="0">
<tr>
<form enctype="multipart/form-data" name="form1" method="post" action="update_ac.php">
<td>
<table width="100%" border="0" cellspacing="1" cellpadding="0">
<tr>
<td>&nbsp;</td>
<td colspan="3"><strong>Update data in mysql</strong> </td>
</tr>
<tr>
<td align="center">&nbsp;</td>
<td align="center">&nbsp;</td>
<td align="center">&nbsp;</td>
<td align="center">&nbsp;</td>
</tr>
<tr>
<td align="center">&nbsp;</td>
<td align="center"><strong>Name</strong></td>
<td align="center"><strong>Main Content</strong></td>
<td align="center"><strong>Image Locatoin</strong></td>
</tr>
<tr>
<td>&nbsp;</td>
<td align="center"><input name="name" type="text" id="name" value="<? echo $rows['name']; ?>"></td>
<td align="center"><input name="mainContent" type="text" id="mainContent" value="<? echo $rows['mainContent']; ?>" size="15"></td>
<td align="center"><input name="photo" type="file" id="photo">
</tr>
<tr>
<td>&nbsp;</td>
<td><input name="id" type="hidden" id="id" value="<? echo $rows['id']; ?>"></td>
<td align="center"><input type="submit" name="Submit" value="Submit"></td>
<td>&nbsp;</td>
</tr>
</table>
</td>
</form>
</tr>
</table>

<?
mysql_close();
 ?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Jonathon Legg - University Year Book - Edit Page</title>
</head>

<body>
<h2>Edit Page (<?php echo ("$adminid"); ?>)</h2>
<a href="test.php"><img src="backtopanel.jpg" alt="back to panel" width="454" height="85" border="0" longdesc="back to panel" /></a>
</body>
</html>

这是 update_acknowledge.php

<?php
session_start();
$UserName = $_SESSION['UserName'];
require("checkLoginSession.php");
include "common.php";
DBConnect();
$Link = mysql_connect($Host, $User, $Password);

//connection removed for stack overflow //
$tbl_name="admin"; // Table name 

$target = "images/"; 
$target = $target . basename( $_FILES['photo']['name']); 
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

// update data in mysql database 
$_POST = array_map("mysql_real_escape_string", $_POST); 
$firstName = $_POST["name"];
$mainText = $_POST["mainContent"];
$pic=($_FILES['photo']['name']); 


//



$sql="UPDATE admin SET name='$firstName', mainContent='$mainText', photo='$pic' WHERE id='$id'";

//

if (mysql_db_query ($DBName, $sql, $Link)){
print ("A record was created <br><a href=index.php> return to index </a>\n");


 //Writes the photo to the server 
 if(move_uploaded_file($_FILES['photo']['tmp_name'], $target)) 
 { 

 //Tells you if its all ok 
 echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory"; 
 } 
 else { 

 //Gives and error if its not 
 echo "Sorry, there was a problem uploading your file."; 
 } 


} else {

print ("Record not created");   
}

mysql_close($Link);
?>

最佳答案

只需用“if”语句换行...

if (!empty($_FILES['photo']['name'])) {
     $pic=mysql_real_escape_string($_FILES['photo']['name']); 

     $sql="UPDATE admin SET name='$firstName', mainContent='$mainText', photo='$pic' WHERE id='$id'";
} else {
     $sql="UPDATE admin SET name='$firstName', mainContent='$mainText' WHERE id='$id'";
}

...当然在 move_uploaded_file block 上也需要类似的 if 语句。

关于php - 使用 PHP 和 MYSQL 上传信息时不需要我的表单字段,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9618344/

相关文章:

mysql - Cassandra 是构建报告系统的最佳方式吗

html - 返回顶部链接的可访问性

php - 需要在该行中选择特定行和特定列并将其与 php 变量进行比较

mysql - 使用 JDBI SQL 对象 API 同时检索或创建新行

php - 正在尝试学习 Drupal 和 PHP 的 ASP.NET 开发人员

java - 一对多和多对一映射在数据库列中显示为空

javascript - 如何仅使用 javascript 换入和换出 DIV 中的文本?

javascript - 如何获取 HTML 表格单元格的内容?

php - 警告 : odbc_connect(): SQL error: [Microsoft][ODBC Driver Manager] Data source name not found and no default driver specified

php - 为什么 xdebug 没有出现在 phpinfo() 中