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我有一個註冊表單,用戶可以通過複選框選擇多個選項。但是現在當我嘗試將選定的數據插入到數據庫中時,它不起作用。下面是一些代碼:如何從PHP中的MYSQL數據庫的下拉列表中插入選中的複選框數據
HTML代碼:
<div class="col-md-6">
<div class="form-group">
<label for="decisions3">Skills</label>
<select name="langOpt2[]" id="langOpt2" multiple="multiple" class="form-control" required data-validation-required-messge="This field is required">
<?php $selectskill = 'select * from skills where status=1';
$dataskill = mysql_query($selectskill);
while($resultskill = mysql_fetch_object($dataskill))
{?>
<option value="<?=$resultskill->skill_name?>"<?php if($result001->skill_name==$resultskill->skill_name){?> selected="selected"<?php } ?> >
<?=$resultskill->skill_name?>
</option>
<?php }
?>
</select>
</div>
的JS
<script>
$('#langOpt').multiselect({
columns: 1,
placeholder: 'Select Languages'
});
$('#langOpt2').multiselect({
columns: 1,
placeholder: 'Select Languages',
search: true
});
$('#langOpt3').multiselect({
columns: 1,
placeholder: 'Select Languages',
search: true,
selectAll: true
});
$('#langOptgroup').multiselect({
columns: 4,
placeholder: 'Select Languages',
search: true,
selectAll: true
});
</script>
,而PHP的: 此代碼將數據插入到數據庫中。我使用舊的PHP版本插入數據。只是爲了練習。
<?php
if(isset($_POST['submit'])){
include("test/admin/includes/db.php");
$skills=$_POST['langOpt2'];
foreach($_POST['langOpt2'] as $skills){
mysql_query("insert into `job_seeker_reg` (`j_skills`) values('','$skills')");
}}
?>
什麼是在這部分的錯誤?出現了什麼錯誤。 –