新手,轻拍. 各位大哥大姐们帮我看一下这段php代码问题出在哪

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Release: 2016-06-23 14:02:30
Original
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Html Code:






 Rep Num: 



 Rep Num: 
Last Name:  First Name: 
Street:  City:  
State:  Zip: 
Commission:  Rate: 



 Rep Num: 



 Rep Num: 
Last Name:  First Name: 
Street:  City:  
State:  Zip: 
Commission:  Rate: 







[/code]


PHP code:
<html><head><title>Sales Rep Maintenance</title></head><?php// Address error handlingini_set('display_errors',1);error_reporting(E_ALL & ~E_NOTICE);// Attempt to Connectif ($connection = @mysql_connect ('localhost', '***', '***')){	print '<p>Successfully connected to MySQL.</p>';	//mysql_close(); // close the connection}else {	die('<p>Could not connect to MySQL because: <b>' .mysql_error() .	'</b></p>');}if (@mysql_select_db("PREMIERE", $connection)){	print '<p>The PREMIERE database has been selected.</p>';}else {	die('<p>Cound not select the PREMIERE database because: <b>' .mysql_error().'</b></p>');}if(!$_REQUEST['submit']){	generate_form();}else if($_REQUEST['submit'] == "Search"){	search_rep();}else if($_REQUEST['submit'] == "Add"){	add_rep();}else if($_REQUEST['submit'] == "Delete"){	delete_rep();}else if($_REQUEST['submit'] == "Update"){	update_rep();}function generate_form(){}function search_rep(){// Define the rep Number php variable name$rep = $_POST['rep_num'];print "The rep num chosen was $rep<br/>";// Define the query$query = "SELECT * FROM REP WHERE CUSTOMER_NUM = '$rep'";// Output the resulting query tableif ($r = mysql_query($query)){	while ($row = mysql_fetch_array($r))	{		print "<p>{$row['REP_NUM']}<br/>{$row['LAST_NAME']}<br/>{$row['FIRST_NAME']}<br/>		{$row['STREET']}<br/>{$row['CITY']}<br/>{$row['STATE']}<br/>{$row['ZIP']}<br/>		{$row['COMMISSION']}<br/>{$row['RATE']}<br/></p>\n";	}}}function add_rep(){}function delete_rep(){}function update_rep(){}?></html>
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可以连接到mySQL服务器上, 但是不能显示结果...


回复讨论(解决方案)

error_reporting(E_ALL & ~E_NOTICE);
改作
error_reporting(E_ALL);
可能就知道了

把 $row 的关联键改成小写看看

error_reporting(E_ALL & ~E_NOTICE);
改作
error_reporting(E_ALL);
可能就知道了

把 $row 的关联键改成小写看看

我试了, 错误提示为:
Undefined index: submit ... on line 28
Undefined index: submit ... on line 33

28行是 if(!$_REQUEST['submit']) 接受的是表单提交的数据
但你的表单中并没有名为 submit 的控件

楼上正解,给type为submit的input加name=submit

28行是 if(!$_REQUEST['submit']) 接受的是表单提交的数据
但你的表单中并没有名为 submit 的控件

那应该怎么改呢?
html文件:

 Rep Num: 

难道不是将input type设置成为"submit"就可以了吗?

加一个隐藏的  input






改为





一般的情况下,input type设置成为"submit"就可以了。但未命名的提交按钮的值是不会被提交的
但是在你的应用中,是靠提交的值按钮来决定操作方式的,所以一定要命名






改为

Thanks a lot!!!

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