dpbz14739 2019-01-28 11:31
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如何在表单标记中包含依赖下拉列表?

I have a working dependent-dropdown list that works smoothly when it's not placed inside a form tag. As soon as I enclose it in a form tag, it stops working. How can I include this code in a form tag?

I have simply tried running the code without using any form tag. It works perfectly fine. To pass the values to the next page I used the form tag and that's when I was getting a blank response(the drop-down list wasn't loading).

index.php

<?php
require_once("dbcontroller.php");
$query ="SELECT * FROM flat";
$results = mysqli_query($conn,$query); 
?>
<html>
<head>
<title></title>

<script src="https://code.jquery.com/jquery-2.1.1.min.js" type="text/javascript"></script>
<script>
function floor(val) {
    $.ajax({
    type: "POST",
    url: "get_info.php",
    data:'floor='+val,
    success: function(data){
        $("#floor").html(data);
    }
    });
}

function flatno(val1) {
    $.ajax({
    type: "POST",
    url: "get_info.php",
    data:'flatno='+val1,
    success: function(data){
        $("#flatno").html(data);
    }
    });
}


</script>
</head>
<body>


<form method="post" action="a.php"><br>


Select Wing:
<select name="wing" onChange="floor(this.value);">
<option value="">---------</option>
<?php
foreach($results as $row) {
?>
<option value="<?php echo $row["wing"]; ?>"><?php echo $row["wing"]; ?>                
</option>
<?php
}
?>
</select>

<br><br>
Select Floor:
<select name="floor" id="floor" onChange="flatno(this.value);">
<option value="">---------</option>
</select>


<br><br>
Select Flat No:
<select name="flatno" id="flatno">
<option value="">---------</option>
</select>

<input type="submit">

</form>

</body>
</html>    

get_info.php

<?php
require_once("dbcontroller.php");

if(!empty($_POST["floor"])) {
$query ="SELECT * FROM flat WHERE wing = '".$_POST["floor"]."'";
$results = mysqli_query($conn,$query); 
?>
    <option value="">Select Floor</option>
<?php
    foreach($results as $row) {
?>
    <option value="<?php echo $row["floor"]; ?>"><?php echo $row["floor"];         
?></option>
<?php
    }
}
?>

<?php
if(!empty($_POST["flatno"])) {
    $query ="SELECT * FROM flat WHERE floor = '".$_POST["flatno"]."'";
    $results = mysqli_query($conn,$query);
?>
    <option value="">Select Flat no</option>
<?php
    foreach($results as $row) {
?>
    <option value="<?php echo $row["flatno"]; ?>"><?php echo     
$row["flatno"]; ?></option>
<?php
    }
}
?>

a.php

<?php

include_once 'dbcontroller.php';

$wing =mysqli_real_escape_string($conn,$_POST['wing']);  
$floor =mysqli_real_escape_string($conn,$_POST['floor']);  
$flatno=mysqli_real_escape_string($conn,$_POST['flatno']); 

echo "<h2>Flat No : " . $flatno . "</h2>";
echo "<h2>Email : " . $floor . "</h2>";
echo "<h2>Email : " . $wing . "</h2>";
?>

I expected all the 3 dropdown list to be functional, but only the 1st dropdown list loads and the other 2 are blank.

  • 写回答

1条回答 默认 最新

  • dpp66953 2019-01-28 12:47
    关注

    you must use different name for your functions for example floor1() and flatno1() because last names( floor() and flatno() ) used for id name of another elements,

    in html codes:

    Change onchange="floor()" to onchange="floor1()"

    Change onchange="flatno()" to onchange="flatno1()"

    in javascript codes:

    Change "function floor()" to "function floor1()"

    Change "function flatno()" to "function flatno1()"

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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