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[英]How to get selected value in one select option and based on that, fetch data from MySQL table to show another select option in the same form
[英]How to get data from database to one select option to another option in PHP MySQL?
我有以下代碼:
<html>
<body>
<form method="post" action="">
Category: <select name="category">
<option>Choose Category</option>
<?php
include("connect.php");
$select="SELECT * FROM category";
$result=mysqli_query($link,$select) or die (mysqli_error($link));
while($row=mysqli_fetch_array($result))
{
echo "<option value='$row[category]'>".$row['category'];
}
?>
</select>
<br>
Subcategory:<select name="category">
<?php
include("connect.php");
$category=@$_POST['category'];
if($category=="Friuts")
{
$select="SELECT * FROM subcategory WHERE $category='$category'";
$result=mysqli_query($link,$select) or die (mysqli_error($link));
while($row=mysqli_fetch_array($result))
{
echo "<option value='echo $row[subcategory]'>".$row['subcategory'];
}
}
?>
</select>
<br>
<input type="submit" value="Open" name="submit">
</form>
</body>
</html>
我有兩個表,表一名稱類別和表二名稱子類別,當我從表類別中選擇一個項目選項時,我想要兩個表,從選擇選項中的表子類別中獲取數據如下圖所示:
要幫您實現所需,您需要:
if( isset( $_POST['category'] ) ) {}
。如果是,請在子類別表上運行查詢並檢查cat_id = $ _POST ['category'] 或簡而言之:
echo "<option value='$row[cat_id]'>".$row['category'];
$select='SELECT * FROM subcategory WHERE cat_id="' . mysqli_real_escape_string($link, $category ).'"';
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