Php 单击时显示基于产品的页面
我有一个页面,其中显示了几个图像,例如,单击milo图像,它会转到一个页面,其中显示了milo的所有图像,但当我单击其他页面时,例如,tea,它也显示了milo的图像。我需要根据单击的图像显示页面Php 单击时显示基于产品的页面,php,mysqli,Php,Mysqli,我有一个页面,其中显示了几个图像,例如,单击milo图像,它会转到一个页面,其中显示了milo的所有图像,但当我单击其他页面时,例如,tea,它也显示了milo的图像。我需要根据单击的图像显示页面 <?php $query = "SELECT * FROM tbl_product ORDER BY id ASC"; //order by ID ascending order $result = mysqli_query($connect, $quer
<?php
$query = "SELECT * FROM tbl_product ORDER BY id ASC"; //order by ID ascending order
$result = mysqli_query($connect, $query);
while($row = mysqli_fetch_array($result))
{
?>
<div id="products" class="productsContainer">
<div class="responsive"> <!-- OUTER BOX OF PRODUCT -->
<div class="gallery"> <!-- INNER BOX FOR PRODUCT -->
<a href="box.php"><img src="images/<?php echo $row["image"]; ?>" class="img-responsive" /></a><br /> <!-- IMAGE OF PRODUCTS -->
<h4 class="text-info"><?php echo $row["name"]; ?></h4> <!-- NAME OF PRODUCT -->
<h4 class="text-danger">$ <?php echo $row["price"]; ?></h4> <!-- PRODUCT PRICE -->
<input type="text" name="quantity" id="quantity<?php echo $row["id"]; ?>" class="form-control" value="1" /> <!-- QUANTITY PRODUCT -->
<input type="hidden" name="hidden_name" id="name<?php echo $row["id"]; ?>" value="<?php echo $row["name"]; ?>" /> <!-- NOT SHOWN -->
<input type="hidden" name="hidden_price" id="price<?php echo $row["id"]; ?>" value="<?php echo $row["price"]; ?>" /> <!-- NOT SHOWN -->
<input type="button" name="add_to_cart" id="<?php echo $row["id"]; ?>" style="margin-top:5px;" class="btn btn-warning form-control add_to_cart" value="Add to Cart" /> <!-- ADD TO CART BUTTON -->
</div>
</div>
</div>
<?php
}
?>
$
将产品Id传入box.php(我相信box.php包含图像细节或产品细节的代码)
例如:
在php中,使用get获取id变量
box.php
<?php
$id = $_GET['id'];
?>
我不知道您知道多少Ajax/Javascript,但在本例中,您可以使用这些代码为您的脚本显示基于其id的图像,因为用户单击image其中id=?
通过Ajax返回返回值:
<script>
function getProduct(val) {
$.ajax({
type: "POST",
url: "box.php",
data:'id='+val,
success: function(data){
$("#product-echo").html(data);
}
});
}
</script>
<?php
$query = "SELECT * FROM tbl_product ORDER BY id ASC"; //order by ID ascending order
$result = mysqli_query($connect, $query);
while($row = mysqli_fetch_array($result))
{
?>
<div id="products" class="productsContainer">
<div class="responsive"> <!-- OUTER BOX OF PRODUCT -->
<div class="gallery"> <!-- INNER BOX FOR PRODUCT -->
<a href="javascript:void(0)" onclick="getProduct('<?php echo $row['id']; ?>')"><img src="images/<?php echo $row["image"]; ?>" class="img-responsive" /></a><br /> <!-- IMAGE OF PRODUCTS -->
<h4 class="text-info"><?php echo $row["name"]; ?></h4> <!-- NAME OF PRODUCT -->
<h4 class="text-danger">$ <?php echo $row["price"]; ?></h4> <!-- PRODUCT PRICE -->
<input type="text" name="quantity" id="quantity<?php echo $row["id"]; ?>" class="form-control" value="1" /> <!-- QUANTITY PRODUCT -->
<input type="hidden" name="hidden_name" id="name<?php echo $row["id"]; ?>" value="<?php echo $row["name"]; ?>" /> <!-- NOT SHOWN -->
<input type="hidden" name="hidden_price" id="price<?php echo $row["id"]; ?>" value="<?php echo $row["price"]; ?>" /> <!-- NOT SHOWN -->
<input type="button" name="add_to_cart" id="<?php echo $row["id"]; ?>" style="margin-top:5px;" class="btn btn-warning form-control add_to_cart" value="Add to Cart" /> <!-- ADD TO CART BUTTON -->
</div>
</div>
</div>
<?php
}
?>
<div id="product-echo"></div>
现在,这很简单,我建议您在处理box.php
端的调用时多花点时间,但是您应该对如何完成这项工作有一个很好的了解
如果您对使用Ajax调用感到不舒服,您可以随时将用户发送到box.php
并启动产品元素。非常有用的答案!我不太擅长ajax,我可以知道val的用途吗?非常感谢,我希望它能帮助您更好地理解它,val
定义要选择的ID。
<?php
if (isset($_POST['id']))
{
$id = $_POST['id'];
$query = "SELECT * FROM tb1_product WHERE id = " . $id;
$result = mysqli_query($connect, $query);
while($row = mysqli_fetch_array($result))
{
// layout content as your heart so desires...
}
}
?>