我有一个小的PHP页面，其中包含两个下拉列表
我需要根据第一个下拉列表中选择的结果填充第二个。这可能吗？换句话说，我需要使用从第一个下拉列表中选择的值，并在用于填充第二个下拉列表的dB查询中使用它（但这应该在选择第一个下拉列表时填充。
如果可以提供任何提示吗？ （您可以假设我能够从dB填充第一个下拉列表）
与db连接
<?php
include ("db_connect.php");
?>
的javascript：
function abc()
{
var val = document.getElementById('customer_type').value;
self.location='add.php?customer_type=' + val ;
}
PHP
<div class="heading">
<h1>Add Products</h1>
</div>
<div class="content">
<form action="" method="post" enctype="multipart/form-data">
<span class="text_line">Select Customer Type:</span>
<select name="customer_type" id="customer_type" onChange="abc()" style="line-height: 30px;
height: 30px; width:18%;">
<option class="tfield"> Customer Type </option>
<?php
$result = mysqli_query($con, "SELECT * FROM cust_type");
while($row = mysqli_fetch_array($result))
{
$cust_name = $row['customer_type'];
echo "<option value='$cust_name'> $cust_name </option>";
}
?>
</select><br />
<span class="text_line">Select Shoes Type:</span>
<select name="shoe_type" style="line-height: 30px;
height: 30px; width:18%;">
<option class="tfield"> Shoes Type </option>
<?php
$result1 = mysqli_query($con, "SELECT * FROM shoes_type");
while($row1 = mysqli_fetch_array($result1))
{
$shoesid = $row1['gender_id'];
$shoes_cat = $row1['shoes_category'];
echo "<option value= '$shoesid'> $shoes_cat </option>";
}
?>
</select><br />
<span class="text_line">Select Image:</span>
<input type="file" name="upload" style="line-height: 30px;
height: 30px; width:18%; color:#fff;"/><br />
<span class="text_line">Price:</span>
<input type="text" name="price" class="tfield"/><br />
<span class="text_line">Description:</span>
<textarea name="descrip" cols="25" class="tfield"></textarea><br />
<div class="button1">
<input type="submit" name="submit" value="Submit" class="sign" />
</div>
</form>
</div>
<?php
if(isset($_POST['submit']))
{
$cust_type = $_POST['customer_type'];
$prod_price = $_POST['price'];
$shoe_img = $_FILES["upload"]["name"];
$prod_descrip = $_POST['descrip'];
$shoeid = $_POST['shoe_type'];
if($cust_type == "" || $prod_price == "" || $shoe_img == "" || $prod_descrip == "")
{
?>
<!-- <div style="color:#FFF; text-align:center; font-size:20px; font-family:Arial, Helvetica, sans-serif; margin-top:2%;">-->
<?php
echo "Please fill all the fields";?>
</div>
<?php
}
else
{
$result2 = mysqli_query($con,
"SELECT * FROM shoes_type WHERE shoes_id = '$shoeid'");
while($row2 = mysqli_fetch_array($result2))
{
$shoesid = $row2['shoes_id'];
$shoetype = $row2['shoes_category'];
}
$sql = mysqli_query($con, "INSERT INTO
product(shoes_id,customer_type,shoe_type,shoe_price,image,description)
VALUES('$shoesid','$cust_type','$shoetype','$prod_price','$shoe_img','$prod_descrip')");
?>
<!--<div style="color:#FFF; text-align:center; font-size:20px; font-family:Arial, Helvetica, sans-serif; margin-top:5%;">-->
<?php
echo "Data inserted !!!";
?>
<?php
//echo "<meta http-equiv='Refresh' content='1'; URL='index.php?id=addproduct'>";
?>
</div>
<?php
}//else loop ends
}//if loop ends
?>
我在ajax和jquery方面经验不足。请在javascript和php中提出解决方案
答案 0 :(得分：0)
我不清楚你的 shoes_type 表中是否有customer_type。但还是试试这个，
与db连接
<?php
include ("db_connect.php");
?>
<强>的javascript：
function abc(){
var val = document.getElementById('customer_type').value;
$.post("getSecondDropDown.php",{ gender_id:val}, function( data ) {
$("#shoe_type").html(data);
});
}
您的主要 PHP
<div class="heading">
<h1>Add Products</h1>
</div>
<div class="content">
<form action="" method="post" enctype="multipart/form-data">
<span class="text_line">Select Customer Type:</span>
<select name="customer_type" id="customer_type" onChange="abc()" style="line-height: 30px;
height: 30px; width:18%;">
<option class="tfield"> Customer Type </option>
<?php
$result = mysqli_query($con, "SELECT * FROM cust_type");
while($row = mysqli_fetch_array($result))
{
$cust_name = $row['customer_type'];
$gender_id = $row['gender_id'];
echo "<option value='$gender_id '> $cust_name </option>";
}
?>
</select><br />
<span class="text_line">Select Shoes Type:</span>
<select name="shoe_type" id="shoe_type" style="line-height: 30px;
height: 30px; width:18%;">
</select><br />
<span class="text_line">Select Image:</span>
<input type="file" name="upload" style="line-height: 30px;
height: 30px; width:18%; color:#fff;"/><br />
<span class="text_line">Price:</span>
<input type="text" name="price" class="tfield"/><br />
<span class="text_line">Description:</span>
<textarea name="descrip" cols="25" class="tfield"></textarea><br />
<div class="button1">
<input type="submit" name="submit" value="Submit" class="sign" />
</div>
</form>
</div>
<?php
if(isset($_POST['submit']))
{
$cust_type = $_POST['customer_type'];
$prod_price = $_POST['price'];
$shoe_img = $_FILES["upload"]["name"];
$prod_descrip = $_POST['descrip'];
$shoeid = $_POST['shoe_type'];
if($cust_type == "" || $prod_price == "" || $shoe_img == "" || $prod_descrip == "")
{
?>
<!-- <div style="color:#FFF; text-align:center; font-size:20px; font-family:Arial, Helvetica, sans-serif; margin-top:2%;">-->
<?php
echo "Please fill all the fields";?>
</div>
<?php
}
else
{
$result2 = mysqli_query($con,
"SELECT * FROM shoes_type WHERE shoes_id = '$shoeid'");
while($row2 = mysqli_fetch_array($result2))
{
$shoesid = $row2['shoes_id'];
$shoetype = $row2['shoes_category'];
}
$sql = mysqli_query($con, "INSERT INTO
product(shoes_id,customer_type,shoe_type,shoe_price,image,description)
VALUES('$shoesid','$cust_type','$shoetype','$prod_price','$shoe_img','$prod_descrip')");
?>
<!--<div style="color:#FFF; text-align:center; font-size:20px; font-family:Arial, Helvetica, sans-serif; margin-top:5%;">-->
<?php
echo "Data inserted !!!";
?>
<?php
//echo "<meta http-equiv='Refresh' content='1'; URL='index.php?id=addproduct'>";
?>
</div>
<?php
}//else loop ends
}//if loop ends
?>
这将是您的新PHP文件，它将获取第二个选择框中的所有值。
<强> getSecondDropDown.php
<?php
$gender_id =$_POST['gender_id '];
$option="";
$result1 = mysqli_query($con, "SELECT * FROM shoes_type where gender_id ='$gender_id'");
while($row1 = mysqli_fetch_array($result1))
{
$shoesid = $row1['gender_id'];
$shoes_cat = $row1['shoes_category'];
$option.= "<option value= '$shoesid'> $shoes_cat </option>";
}
echo $option;
?>
还有一件事，你需要添加jQuery库才能工作。