php：处理选择标记
[英]php:process the select tag
问题描述
this code run but when test it .. the data not correct i select data from database and i make it in array ad show it id by select tag but when select any id and submit .. example i select 5 and click on submit the record will delete is 2 not 5 
此代码运行，但是当测试它..数据不正确时，我从数据库中选择数据，并在数组广告中将其显示为通过选择标签显示ID，但是当选择任何ID并提交时。例如，我选择5并单击提交记录将删除的是2而不是5
<?php
require_once "config.php";
$qid="select id from info_user";
$arrayid=array();
$result=mysql_query($qid);
while($res=mysql_fetch_array($result)){
$arrayid[]=$res['id'];
}
var_dump($arrayid);
if(isset($_POST['sub'])){
$id=$_POST['id'];
$q="delete from info_user where id=$id ";
$qq=mysql_query($q);
if($qq){
echo "you delete record ";
}
else{
echo "error in deleting";
}}
?>
<html>
<head>
<title>delete</title>
</head>
<form action="delete.php" method="post">
<select name="id"><?php for($i=0;$i<count($arrayid);$i++){?>
<option value="<?php echo $i;?>"><?php echo $arrayid[$i];} ?></option></select> <br />
<input type="submit" name="sub" />
</form>
</html>
3 个解决方案
解决方案1
0 已采纳 2014-09-30 12:35:39
I think that problem is in value for the options. 我认为该问题对于期权具有价值。
Try changing option line to 尝试将选项行更改为
<option value="<?php echo $arrayid[$i];?>"><?php echo $arrayid[$i];} ?></option></select> <br />
解决方案2
0 2014-09-30 12:38:55
Your <option> markup is wrong. 您的<option>标记是错误的。 Make your markup like this; 使您的标记像这样；
<select name="id">
<?php for ($i = 0; $i < count($arrayid); $i++) { ?>
<option value="<?php echo $i; ?>"><?php echo $arrayid[$i]; ?></option>
} ?>
</select>
Tip: You'll find debugging easier if you indent your code. 提示：如果缩进代码，您会发现调试会更容易。 IDE's generally have this option. IDE通常具有此选项。 NetBeans is my favourite. NetBeans是我的最爱。
解决方案3
0 2014-10-01 13:25:32
Instead of sending the ids, you're sending an array key associated to the id. 而不是发送ID，而是发送与ID关联的数组键。
For instance, assume that: $arrayid = array(10, 11, 12); 例如，假定： $arrayid = array(10, 11, 12); . 。
This is equivalent to: $arrayid = array(0 => 10, 1 => 11, 2 => 12); 这等效于： $arrayid = array(0 => 10, 1 => 11, 2 => 12); . 。
You'll see the options 10, 11 and 12, but you'll send either 0, 11 or 12, because that's what you're setting to the options values: 您将看到选项10、11和12，但是将发送0、11或12，因为这是您要为选项值设置的内容：
<select name="id">
<option value="0">10</option>
<option value="1">11</option>
<option value="2">12</option>
</select>
If you select the option "11", the SQL statement to delete an entry will be: 如果选择选项“ 11”，则删除条目的SQL语句将为：
delete from info_user where id=1
I did not test this code and I'm assuming the ids are integers, but try it: 我没有测试此代码，并且我假设id是整数，但是尝试一下：
<?php
require_once "config.php";
$qid = "select id from info_user";
$arrayid = array();
$result = mysql_query($qid);
while( $res = mysql_fetch_array($result) ){
$arrayid[] = $res['id'];
}
if( isset($_POST['sub']) ){
$id = (int)$_POST['id']; // make sure it's an integer to avoid SQL injections
$q = "delete from info_user where id=$id";
$qq = mysql_query($q);
if( $qq ) {
$error = "you delete record ";
} else {
$error = "error in deleting";
}
}
?>
<!DOCTYPE html>
<html>
<head>
<title>delete</title>
</head>
<body>
<?php
if( isset($error) ) :
echo '<p>', htmlspecialchars($error), '</p>';
elseif( !empty($arrayid) ) :
var_dump($arrayid);
endif;
?>
<form action="delete.php" method="post">
<select name="id">
<?php foreach($arrayid as $id): ?>
<option value="<?php echo (int)$id;?>">
<?php echo (int)$id; ?>
<?php endforeach; ?>
</option>
</select>
<br />
<input type="submit" name="sub" />
</form>
</body>
</html>
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