我正在尝试使用先前创建的动态复选框从mysql表中提取数据,但是看起来我的搜索只显示最多2条记录,如果我选择了3个以上的复选框,它将不会返回任何内容,所以我想知道是否有人可以帮我弄清楚怎么做。
以下是我的代码,非常感谢您的帮助:
$warehouse = @$_POST['wh'];
switch($button){
case 'Submit':
if(@$_POST['wh']){
$warehouse = @$_POST['wh'];
$length = count($warehouse);
for ($i = 0; $i < $length; $i++){
//echo '<br>'.$warehouse[$i];
}
$consult = "SELECT * FROM contact_info WHERE ";
for ($i = 0; $i < $length; $i++){
$consult = $consult . "warehouse='$warehouse[$i]'";
if(!$i+1 == $length){
$consult = $consult . " OR ";
}
}
echo '<br>'.$length.'<br>';
print_r ($consult);
$response = mysqli_query($connection, $consult);
if($response){
$registry = mysqli_affected_rows($connection);
if($registry > 1){
echo '<br><table width="800" align="center" border="2" cellspacing="1" cellpadding="1">
<form action="index.php" method="post" >
<tr>
<td align="center"><strong>Warehouse</strong></td>
<td align="center"><strong>Name</strong></td>
<td align="center"><strong>Lastname</strong></td>
<td align="center"><strong>Phone</strong></td>
<td align="center"><strong>I-net</strong></td>
<td align="center"><strong>E-mail</strong></td>
</tr>';
while($registry = mysqli_fetch_array($response, MYSQLI_ASSOC)){
echo '<form action="index.php" method=post >
<tr>
<td div align="center">'.$registry['warehouse'].'
<td div align="center">'.$registry['name'].'
<td div align="center">'.$registry['lastname'].'
<td div align="center">'.$registry['phone'].'
<td div align="center">'.$registry['inet'].'
<td div align="center">'.$registry['email'].'
</tr>';
}
echo '</form>';
echo '
</form>
</table>';