PHP多重功能更新不更新，并得到第1行空白 [英] PHP Multiple Update function not updating and gets the 1st row blank
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问题描述
首先，我用PHP Dreamweaver将与phpMyAdmin数据库的程序。这一方案得到我的记录在第一排的空白，当我提交按钮，然后在一个行，其余的记录仍然存在这是第二排及以下，但在第1行中的记录仍然在数据库中，它为什么要发生？任何想法为我解决这个程序，它也不会更新我的纪录。这让我感到恶心，我需要帮助，请！ :(
TNX ...
这是code：
＆LT; PHP require_once（'连接/ tlsc_conn.php'）;
mysql_select_db（$ database_tlsc_conn，$ tlsc_conn）;
$ query_Recordset1 =SELECT * FROM tbl_name表;
$ Recordset1 =的mysql_query（$ query_Recordset1，$ tlsc_conn）或死亡（mysql_error（））;
$ row_Recordset1 = mysql_fetch_assoc（$ Recordset1）;
$ totalRows_Recordset1 = mysql_num_rows（$ Recordset1）; 如果（使用isset（$ _ POST ['提交']））{
$数= COUNT（$ _ POST ['身份证']）;
$提交= $ _GET ['身份证']; 为（$ I = 0; $ I＆LT; $计数; $ I ++）{
$ SQL1 =UPDATE $ tbl_name表SET NAME ='$名称[$ i]，姓氏='$姓氏[$ i]'，电子邮件= $电子邮件[$ i]'WHERE ID ='$ ID [$ i] ';
$ row_Recordset1 =的mysql_query（$ SQL1）;
} 如果（$ row_Recordset1）{
标题（位置：lulu.php）;
出口;
}
}
？＆GT;
！＆LT; DOCTYPE HTML PUBLIC - // W3C // DTD XHTML 1.0过渡// ENhttp://www.w3.org /TR/xhtml1/DTD/xhtml1-transitional.dtd\">
＆LT; HTML的xmlns =http://www.w3.org/1999/xhtml＆GT;
＆LT; HEAD＆GT;
＆LT; META HTTP-EQUIV =Content-Type的CONTENT =text / html的;字符集= ISO-8859-1/＆GT;
＆LT;标题＆GT;无标题文档＆LT; /标题＆GT;
＆LT; /头＆GT;
＆LT;身体GT;
＆LT;表格名称=窗体2的方法=邮报行动=＆GT;
＆LT;表格的宽度=634BORDER =1＆GT;
＆所述; TR＆GT;
＆LT; TD＆GT; ID＆LT; / TD＆GT;
＆LT; TD＆GT;名称和LT; / TD＆GT;
＆LT; TD＆GT;姓氏＆LT; / TD＆GT;
＆LT; TD＆GT;电子邮件＆LT; / TD＆GT;
＆LT; / TR＆GT;
＆LT; PHP做{＆GT？;
＆所述; TR＆GT;
＆LT; TD＆GT;＆LT; PHP的$ id [] = $ row_Recordset1 ['身份证']; ？＆GT;＆LT; PHP的echo $ row_Recordset1 ['身份证']; ？＆GT;
＆LT;输入名称=ID []类型=隐藏值=＆LT; PHP的echo $ row_Recordset1 ['身份证'];？＆gt;中/＆GT;
＆LT; / TD＆GT;
＆所述; TD＆GT;
＆LT;输入名称=名称[]类型=文本VALUE =＆LT; PHP的echo $ row_Recordset1 ['名'];？＆GT;＆GT;
＆LT; / TD＆GT;
＆所述; TD＆GT;
＆LT;输入名称=姓氏[]类型=文本VALUE =＆LT; PHP的echo $ row_Recordset1 ['姓氏'];？＆GT;＆GT;
＆LT; / TD＆GT;
＆所述; TD＆GT;
＆LT;输入名字=电子邮件[]类型=文本VALUE =＆LT; PHP的echo $ row_Recordset1 [电子邮件];？＆GT;＆GT; ＆LT; / TD＆GT;
＆LT; / TR＆GT;
＆LT; PHP}而（$ row_Recordset1 = mysql_fetch_assoc（$ Recordset1））; ？＆GT;
＆LT; /表＆gt;
＆所述p为H.;
＆LT;输入类型=提交名称=提交值=提交/＆GT;
＆所述; / P＆GT;
＆LT; /表及GT;
＆所述p为H.;
＆所述; / P＆GT;
＆LT; /身体GT;
＆LT; / HTML＆GT;
解决方案
您需要把：
$ NAME = array_map（'mysql_real_escape_string'，$ _ POST ['名']）;
$姓氏= array_map（'mysql_real_escape_string'，$ _ POST ['姓氏']）;
$电子邮件= array_map（'mysql_real_escape_string'，$ _ POST ['邮件']）;
$ ID = array_map（'mysql_real_escape_string'，$ _ POST ['身份证']）;
在你的为循环。您使用的这些变量，但你永远充满他们从表单输入。
First of all, I use PHP Dreamweaver to make a program with phpMyAdmin Database. This program gets my records in a 1st row blank when I submit the button, and the remaining records in a rows are still there which is 2nd row and below, but the records in a 1st row is still on the database, why should it happen? any idea for me to solve this program, it also won't update my record. this makes me sick, i need help please! :( tnx...
this is the code:
<?php require_once('Connections/tlsc_conn.php');
mysql_select_db($database_tlsc_conn, $tlsc_conn);
$query_Recordset1 = "SELECT * FROM tbl_name";
$Recordset1 = mysql_query($query_Recordset1, $tlsc_conn) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1);
if(isset($_POST['submit'])) {
$count = count($_POST['id']);
$submit = $_GET['id'];
for($i=0;$i<$count;$i++){
$sql1="UPDATE $tbl_name SET name='$name[$i]', lastname='$lastname[$i]', email='$email[$i]' WHERE id='$id[$i]'";
$row_Recordset1=mysql_query($sql1);
}
if($row_Recordset1){
header("location:lulu.php");
exit;
}
}
?>
<!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=iso-8859-1" />
<title>Untitled Document</title>
</head>
<body>
<form name="form2" method="post" action="">
<table width="634" border="1">
<tr>
<td>id</td>
<td>name</td>
<td>lastname</td>
<td>email</td>
</tr>
<?php do { ?>
<tr>
<td><?php $id[]=$row_Recordset1['id']; ?><?php echo $row_Recordset1['id']; ?>
<input name="id[]" type="hidden" value="<?php echo $row_Recordset1['id']; ?>" />
</td>
<td>
<input name="name[]" type="text" value="<?php echo $row_Recordset1['name']; ?>">
</td>
<td>
<input name="lastname[]" type="text" value="<?php echo $row_Recordset1['lastname']; ?>">
</td>
<td>
<input name="email[]" type="text" value="<?php echo $row_Recordset1['email']; ?>"> </td>
</tr>
<?php } while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)); ?>
</table>
<p>
<input type="submit" name="submit" value="Submit" />
</p>
</form>
<p>
</p>
</body>
</html>
解决方案
You need to put:
$name = array_map('mysql_real_escape_string', $_POST['name']);
$lastname = array_map('mysql_real_escape_string', $_POST['lastname']);
$email = array_map('mysql_real_escape_string', $_POST['email']);
$id = array_map('mysql_real_escape_string', $_POST['id']);
before your for loop. You're using these variables but you never filled them in from the form inputs.
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