获得多个ID,同时POST多个选择选项值到下一个表格
问题描述:
如何获得多个ID,同时POST多个选择选项值到下一个表格?我只从数组中获得第一个选择ID。你们可以向我提出任何想法吗?获得多个ID,同时POST多个选择选项值到下一个表格
这里是我的代码,当选择的值。
<tr>
<label>Auditor: </label>
<select class="form-control" name="auditor[]" multiple="multiple" >
<?php
$result = $db->query("SELECT * FROM auditor");
while($row = mysqli_fetch_array($result))
{
echo '<option value="'.$row["auditor_name"].'">'.$row["auditor_name"].'</option>';
}
echo "</select>";
?>
</tr>
这是另一个代码,而POST到下一页。
$myselected = $_POST["auditor"];
if(count($myselected)>1){
$auditor = implode ("','",$myselected);
}else{
$auditor =$myselected;
}
$query10 = "SELECT * FROM auditor WHERE auditor_name IN ('$auditor') ";
$result10 = $db->query($query10);
$row10 = $result10->fetch_array();
?>
<form action="audit_action/audit_action.php" role="form" method="post" name="auditformdetails" onsubmit="return(validate());">
<table width='100%' border='0' class="table">
<tr>
<td colspan=6>Audit details</td>
<td colspan=6>Outlet details</td>
</tr>
<tr>
<td><b>Auditor:</b></td>
<td colspan='5'>
**<?php
echo'<input type="hidden" name="auditor_id" value="'.$row10["id"].'">';
foreach ($myselected as $auditor){
echo $auditor."<br>\n";
}
?>**
</td>
答
您无法将字符串与mysql IN子句进行比较。所以,你必须连接你的每个值或查询条件,如下所示。
$myselected = $_POST["auditor"];
$sql_cond = "";
if(count($myselected)>1){
foreach($myselected as $selected){
if($sql_cond != "")
$sql_cond.=" or auditor_name = ".$selected;
else
$sql_cond.=" auditor_name = ".$selected;
}
}else{
$auditor =$myselected;
}
$query10 = "SELECT * FROM auditor WHERE ".$sql_cond;
+0
查询后的$ result语句如何?只需跟着 $ result10 = $ db-> query($ query10); $ row10 = $ result10-> fetch_array(); 是否正确? – Andrew
你能解释一下你想要的输出吗? – sandeepsure
嗨@sandeepsure,输出例如: ROBERT ALAN ALVIN 但是,当后的数据也只显示前ROBERT“ID”;另外两个ALAN和ALVIN id没有显示,当我传递表单时,它只存储第一个ROBERT ID到数据库中。 – Andrew
'print_r($ myselected)'让我知道结果。 – sandeepsure