Php 一个简单的操作,但它给出了一个错误,我想更新每个id
这很简单,但当我按下submit时,我没有得到更新单选按钮的值 foreach语句中出现错误,如:警告:为foreach()提供的参数无效请说明如何在按下submit按钮后获取单选按钮的值, 这里我正在更新状态Php 一个简单的操作,但它给出了一个错误,我想更新每个id,php,Php,这很简单,但当我按下submit时,我没有得到更新单选按钮的值 foreach语句中出现错误,如:警告:为foreach()提供的参数无效请说明如何在按下submit按钮后获取单选按钮的值, 这里我正在更新状态 <?php include("db.php"); ?> <html> <body> <?php $state=$_POST['state']; if(isset($
<?php
include("db.php");
?>
<html>
<body>
<?php
$state=$_POST['state'];
if(isset($_POST['submit']))
{
$result1 = mysql_query("select state,id,status from states ");
while($rr1=mysql_fetch_array($result1))
{
//getting values of radio buttons
$myval=$rr1['id'];
echo $myval;
$val=$_POST['yes.$rr1["id"]'];
echo $val;
$val1=$val.$myval;
$vall=yes.$rr1['id'];
foreach($vall as $values)
{
echo $values;
$update=mysql_query("UPDATE states SET status='". $values."'
WHERE id='$myval' ");
}
}
}
echo "ok";
?>
<form action="" name="form" id="form" method="post" >
<?php
//session_start();
include("db.php");
$result = mysql_query("select state,id,status from states ");
while($info1=mysql_fetch_assoc( $result))
{
echo $info1['city'];
echo "<br>";
/*echo "<br>";
echo "company Name:".$info1['company_name'];
echo "<br>";
echo "salary:".$info1['maxsalary'];
echo "<br>";
//echo $info1['company_name'];*/
?>
<label><?php echo $info1['state']; ?></label>
<input type="radio" <?php if($info1['status']=="yes"){ echo
"checked='checked'"; } ?> name="yes.<?php echo $info1[ 'id']; ?>[]"
value="yes">Yes
<input type="radio" <?php if($info1['status']=="no"){ echo
"checked='checked'"; } ?> name="yes.<?php echo $info1[ 'id']; ?>[]"
value="no">no
<br/>
<?php } ?>
<br />
<input type="submit" name="submit" value="submit" />
</form>
</body>
</html>
我想你需要写信
$vall=yes.$rr1['id'];
to
$vall="yes".$rr1['id'];
谢谢