PHP的更新查询失败my_fetch_array [英] php UPDATE QUERY fail my_fetch_array
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问题描述
我想在表单中更新我的供应商信息。
下面是我的我的code为editsupplier形式
< PHP
$供应商ID = $ _GET ['供应商ID'] = $行['供应商ID'];
//连接并选择数据库
mysql_connect(localhost的根,);
mysql_select_db(supplierdetails);
//运行查询
$ RESULT1 =的mysql_query(SELECT * FROM供应商WHERE供应商id = $供应商ID);
而($行= mysql_fetch_array($ RESULT1)){
$供应商ID = $ _GET ['供应商ID'] = $行['供应商ID'];
$ SupplierName = $行['SupplierName'];
$货币= $行['货币'];
$位置= $行['位置'];
$ ContactNumber = $行['ContactNumber'];
$电子邮件= $行[电子邮件];
}
?>
!< DOCTYPE HTML PUBLIC - // W3C // DTD XHTML 1.0 Strict标准// ENhttp://www.w3.org/TR/xhtml1/DTD/xhtml1- strict.dtd>
< HTML的xmlns =http://www.w3.org/1999/xhtmlXML:LANG =ENLANG =ENGT&;
< HEAD>
< META HTTP-EQUIV =内容类型内容=text / html的;字符集= UTF-8/>
< META NAME =说明内容=/>
< META NAME =关键词内容=/>
< META NAME =作者内容=/>
<链接rel =stylesheet属性类型=文/ CSS的href =style.css文件媒体=屏幕/>
<标题>哈里森·斯平克斯< /标题>
< /头>
<身体GT;
< DIV ID =包装>
< PHP的包括:(包括/ header.php文件'); ?>
< PHP的包括:(包括/ nav.php'); ?>
< DIV ID =内容>
< H3><中心>编辑供应商信息< /中心及GT;< / H3 GT&;
< P>请输入以下所有细节编辑供应商< / P>
< / DIV>
< DIV>
< BR>
< / BR>
<形式的行动=Edit_Supp.php方法=后>
< BR>
< / BR>
<输入类型=隐藏的名字=供应商IDVALUE =< PHP的echo $供应商ID;>中/> 供应商名称:<输入类型=文本名称=SupplierNameVALUE =?< PHP的echo $ SupplierName;>? />
< BR>
< / BR>
货币:<输入类型=文本名称=货币值=< PHP的echo $货币;>有 />
< BR>
< / BR>
位置:其中,INPUT TYPE =文本名称=位置值=< PHP的echo $位置;>有 />
< BR>
< / BR>
联系电话:LT;输入类型=文本名称=ContactNumberVALUE =?< PHP的echo $ ContactNumber;>中/>
< BR>
< / BR>
电子邮件:;?< PHP的echo $电子邮件;>中和LT输入类型=文本名称=电子邮件值= />
< BR>
< / BR>
<输入类型=提交值=编辑供应商信息/> < /表及GT;
< / DIV>
< /身体GT;
< / HTML>
错误是:
注意:未定义的变量:行E:\\ XAMPP \\ htdocs中\\ EditSupForm.php第2行
警告:mysql_fetch_array()预计参数1是资源,布尔在E中给出:\\ XAMPP \\ htdocs中\\ EditSupForm.php第8行
这是在形式背后的code:
< PHP
$ CON = mysql_connect(localhost的根,);
mysql_select_db(supplierdetails);
如果(!$ CON)
{
死亡(无法连接:'mysql_error());
}
//运行查询
$供应商ID = $ _ POST ['供应商ID'] = $行[供应商ID];
$ RESULT1 =的mysql_query(SELECT * FROM供应商WHERE供应商id =$供应商ID。')或死亡(mysql_error());
$行= mysql_fetch_array($ RESULT1);
$供应商ID = $ _ POST ['供应商ID'] = $行[供应商ID];
$ SupplierName = $ _ POST ['SupplierName'];
$货币= $ _ POST ['货币'];
$位置= $ _ POST ['位置'];
$ ContactNumber = $ _ POST ['ContactNumber'];
$电子邮件= $ _ POST [电子邮件];
$供应商ID = $行['供应商ID'];
$查询=更新供应商SET供应商id ='。$供应商ID。',SupplierName ='。$ SupplierName。',货币='。$货币。',位置='。$位置。' ,ContactNumber ='。$ ContactNumber',电子邮件='。$电子邮件'WHERE供应商id ='的$ id。';
$ RESULT1 =的mysql_query($查询);
//检查查询是否成功与否
如果($ RESULT1)
{
回声供应商更新了;
标题(位置:Supplier.php);
}
其他
{
死(查询失败);
}
?>
解决方案
在这两个你使用 $行[供应商ID]你的code的;
前执行MySQL查询,`$行[供应商ID];这就是为什么你越来越错误。
I am trying to update my supplier information within a form
Here is my my code for the editsupplier form
<?php
$SupplierID = $_GET['SupplierID'] = $row['SupplierID'];
//Connect and select a database
mysql_connect ("localhost", "root", "");
mysql_select_db("supplierdetails");
//Run query
$result1 = mysql_query("SELECT * FROM suppliers WHERE SupplierID=$SupplierID");
while($row = mysql_fetch_array($result1)){
$SupplierID = $_GET['SupplierID'] = $row['SupplierID'];
$SupplierName = $row['SupplierName'];
$Currency = $row['Currency'];
$Location = $row['Location'];
$ContactNumber = $row['ContactNumber'];
$Email = $row['Email'];
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1- strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<meta http-equiv="content-type" content="text/html; charset=utf-8" />
<meta name="description" content="" />
<meta name="keywords" content="" />
<meta name="author" content="" />
<link rel="stylesheet" type="text/css" href="style.css" media="screen" />
<title>Harrison Spinks</title>
</head>
<body>
<div id="wrapper">
<?php include('includes/header.php'); ?>
<?php include('includes/nav.php'); ?>
<div id="content">
<h3><center>Edit Supplier Information</center></h3>
<p>Please enter all the following details to edit a supplier</p>
</div>
<div>
<br>
</br>
<form action="Edit_Supp.php" method="post">
<br>
</br>
<input type="hidden" name="SupplierID" value="<?php echo $SupplierID;?>"/>
Supplier Name: <input type="text" name="SupplierName" value="<?php echo $SupplierName ;?>" />
<br>
</br>
Currency: <input type="text" name="Currency" value="<?php echo $Currency ;?>" />
<br>
</br>
Location: <input type="text" name="Location" value="<?php echo $Location ;?>" />
<br>
</br>
Contact Number:<input type="text" name="ContactNumber" value="<?php echo $ContactNumber ;?>" />
<br>
</br>
Email:<input type="text" name="Email" value="<?php echo $Email ;?>" />
<br>
</br>
<input type="submit" value= "Edit Supplier Information"/>
</form>
</div>
</body>
</html>
Errors are:
Notice: Undefined variable: row in E:\xampp\htdocs\EditSupForm.php on line 2
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in E:\xampp\htdocs\EditSupForm.php on line 8
This is the code behind the form:
<?php
$con = mysql_connect("localhost", "root", "");
mysql_select_db("supplierdetails");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
//Run a query
$SupplierID= $_POST['SupplierID'] = $row ["SupplierID"];
$result1 = mysql_query ("SELECT * FROM suppliers WHERE SupplierID= '".$SupplierID."'") or die(mysql_error());
$row = mysql_fetch_array($result1);
$SupplierID = $_POST['SupplierID'] = $row ["SupplierID"];
$SupplierName = $_POST['SupplierName'];
$Currency = $_POST['Currency'];
$Location = $_POST['Location'];
$ContactNumber = $_POST['ContactNumber'];
$Email = $_POST['Email'];
$SupplierID = $row['SupplierID'];
$query = "UPDATE suppliers SET SupplierID = '".$SupplierID."', SupplierName= '".$SupplierName."', Currency = '".$Currency."', Location = '".$Location."', ContactNumber = '".$ContactNumber."', Email = '".$Email."' WHERE SupplierID = '".$id."'";
$result1 = mysql_query($query);
//Check whether the query was successful or not
if($result1)
{
echo "Supplier Updated";
header ("Location: Supplier.php");
}
else
{
die ("Query failed");
}
?>
解决方案
In both of your code you are using $row ["SupplierID"];
before performing mysql query, `$row ["SupplierID"];' thats why your are getting errors.
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