插入是好的，但如果我从数据库中选择这个值，所有值都很好，但单值获取数组问题，我不知道如何解决这个任务。请尽快更新此代码。
这是插入代码.....
<?php
if(isset($_POST['sendmessage'])){
$entermessage = $_POST['teachermessage'];
$checkbox_user = $_POST['usernameallcheckbx'];
$arr = implode(',',$checkbox_user);
//$arr2 = explode(',',$arr);
$insert = mysql_query("INSERT into usermessages(fromteacher,toparent,messages) VALUES('".$_SESSION['username']."','$arr','$entermessage')");
if($insert == 1){
echo "<h1>successful</h1>";
}
else{
echo mysql_error();
}
}
?>
<form method="post">
<div style="float:left; width:450px;"><br/><br/>Message: <br/>
<textarea style="width:400px; height:300px;" name="teachermessage"></textarea><br/>
<input type="submit" value="Send" name="sendmessage" /></div>
<div style="float:left; with:200px; padding-top:55px;">
<table>
<tr>
<?php
$select_query1 = mysql_query("SELECT * FROM register_user WHERE teacher='$teachername'");
while($chckbx=mysql_fetch_array($select_query1))
{
?>
<td><?php echo "<input type='checkbox' name='usernameallcheckbx[]' value=". $chckbx['userid']." />"; ?></td>
</tr>
<tr>
<td><?php echo $chckbx['parent_fname']." ".$chckbx['parent_lname']; ?></td>
</tr>
<?php
}
?>
</table>
</div>
</form>
这是选择代码....
<h2>Messages</h2>
<?php
$select_query3 = mysql_query("SELECT * FROM usermessages");
$fetch= mysql_fetch_array($select_query3);
$data=$fetch['toparent'];
$arr=explode(',',$data);
//$userids=explode(',',$data);
$sessionshow=$_SESSION['userid'];
$userids=in_array("$sessionshow",$arr);
$select_query2 = mysql_query("SELECT * FROM usermessages WHERE toparent in ('$userids')='".$_SESSION['userid']."'");
?>
<table border="0" cellpadding="0" cellspacing="0">
<tr>
<th>Teacher Name</th>
<th>Message</th>
</tr>
<?php
while($fetch_name2=mysql_fetch_array($select_query2))
{
echo "<tr>";
echo "<td>".$fetch_data=$fetch_name2['fromteacher']."</td>";
echo "<td>".$fetch_data=$fetch_name2['messages']."</td>";
echo "</tr>";
}
?>
</table>
我只把这个作为一个答案来显示一些不正确的代码-从你的循环中删除获取数据，因为它不做任何事情
while($fetch_name2=mysql_fetch_array($select_query2))
{
echo "<tr>";
echo "<td>".$fetch_name2['fromteacher']."</td>";
echo "<td>".$fetch_name2['messages']."</td>";
echo "</tr>";
}
更好的是使用HEREDOC:
while($fetch_name2=mysql_fetch_array($select_query2))
{
echo <<<EOF
<tr>
<td>{$fetch_name2['fromteacher']}</td>
<td>{$fetch_name2['messages']}</td>
</tr>
EOF;
}
我真的不知道你想做什么，但也许这是你想要的查询:
SELECT * FROM usermessages WHERE FIND_IN_SET('$sessionshow', toparent)
FIND_IN_SET(str, strlist)在strlist中以逗号分隔的列表中查找与str相等的元素，并返回其在列表中的位置;如果没有找到，则返回0，其计数为false。
我看不出使用$select_query3的任何代码的任何目的。