[英]Dynamically Populating each rows from database using php-Ajax
我有一个从mysql数据库动态填充的表,希望用户选择一个staff number
,该staff number
会自动转到数据库并获取其人员编号。我喜欢10行。 它适用于第一行,但不适用于随后的其他行。
请看一下我遗漏的代码和建议。
谢谢
<tr>
<th nowrap="nowrap">S/N</th>
<th nowrap="nowrap">VNO</th>
<th nowrap="nowrap">Name</th>
<th nowrap="nowrap">Staff No</th>
</tr>
<tr>
<?php
$c=0;
$st =mysqli_query($connection,"SELECT * FROM tab_flt WHERE mainloc='".$_SESSION['location']."' AND status='Active'");
while($r = mysqli_fetch_array($st)){ $c++?>
<td><?php echo $c;?></td>
<td><input type="text" name="flt[]" value="<?php echo $r['fltno'];?>" class="form-control" readonly="readonly" /></td>
<td><select name="opname[]" class="form-control" id="subloc">
<option>Select...</option>
<?php
$fs = getOperators($_SESSION['location']);
while($f = mysqli_fetch_array($fs)){?>
<option value="<?php echo $f['name'];?>"><?php echo $f['name'];?></option>
<?php };?>
</select></td>
<td id="staffno"></td>
</tr>
Ajax方面:
<script type="text/javascript">
$(document).ready(function() {
$("#subloc").change(function(){
var sname = $("#subloc option:selected").val();
$.ajax({
type:"POST",
url:"process-opno.php",
data:{opname:sname}
}).done(function(data3){
$("#staffno").html(data3);
});
});
});
</script>
当成功将subloc id选择为staffno id时,以上代码获取第一行。 但是对于其余的行则不执行此操作。 我该怎么做,以便它将识别第二行,第三行等并将相应的人员编号提取到staffno id中。
谢谢。
请尝试以下方法:
PHP部分我已经加入classes
的选择框,选择框AJAX结果
<tr>
<?php
$c=0;
$st =mysqli_query($connection,"SELECT * FROM tab_flt WHERE mainloc='".$_SESSION['location']."' AND status='Active'");
while($r = mysqli_fetch_array($st)){
$c++;
?>
<td><?php echo $c;?></td>
<td><input type="text" name="flt[]" value="<?php echo $r['fltno'];?>" class="form-control" readonly="readonly" /></td>
<td>
<select name="opname[]" class="form-control js-sel-box" data-id="<?php echo $c;?>">
<option>Select...</option>
<?php
$fs = getOperators($_SESSION['location']);
while($f = mysqli_fetch_array($fs)){
?>
<option value="<?php echo $f['name'];?>"><?php echo $f['name'];?></option>
<?php
};
?>
</select>
</td>
<td class="js-sel-box-ajax-result-<?php echo $c;?>"></td>
<?php
}//End While
?>
</tr>
Ajax部分:
<script type="text/javascript">
$(document).ready(function() {
$(".js-sel-box").change(function(){
var sname = $(this).val();
var result_id = $(this).attr("data-id");
$.ajax({
type:"POST",
url:"process-opno.php",
data:{opname:sname}
}).done(function(data3){
$(".js-sel-box-ajax-result-"+result_id).html(data3);
});
});
});
</script>
与唯一ID相同的旧问题,将ID更改为一个类,根据更改后的选择查找所有元素
<script type="text/javascript">
$(document).ready(function() {
$("td select.form-control").change(function(){
var sname = $(this).val();
var el = $(this);
$.ajax({type:"POST",url:"process-opno.php",data:{opname:sname}})
.done(function(data3){
$(this).parent().next("td").html(data3);
});
});
});
</script>
是的,最好使用类标识符而不是id来标识多个DOM元素; 但在您的情况下,只需进行最小的更改就可以使用-
// use
var sname = $(this).val();
//put this line just var sname
var $object =$(this);
// instead of
var sname = $("#subloc option:selected").val();
// this is because- $('#subloc option:selected').val(); always returns the first
//dropdownList/optionlist from the DOM array, you can use `this` to track which DOM has been change recently
// another change in your code $object is $(this) equivalent but will not work inside ajax so you need to create $object as in code above..
// put this line
$($object).parents('tr').find("#staffno").html(data3);
//instead of
$("#staffno").html(data3);
// above line will search parent tr and will look for DOM element with id staffno a
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