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Php 网页无法找到连接变量

转载 作者:行者123 更新时间:2023-11-29 05:54:14 26 4
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我正在开发在线订购系统。我想在服务器加载网页时显示产品详细信息。我的产品详细信息存储在 mysql 数据库中。但问题是当我在本地主机上运行应用程序时出现以下错误..

Notice: Undefined variable: con in C:\wamp64\www\onlineordering\functions\datafetching.php on line 33

Warning: mysqli_query() expects parameter 1 to be mysqli, null given in C:\wamp64\www\onlineordering\functions\datafetching.php on line 33

这是我的connection.php

<?php

$servername = "localhost";
$username = "root";
$password = "";
$db = "ecom1";

// Create connection
$con = mysqli_connect($servername, $username, $password,$db);

// Check connection
if (!$con) {
die("Connection failed: " . mysqli_connect_error());
}


?>

这是 datafecting.php 的代码。

<?php
include "db/connection.php";

?>

<?php


function getrecords($exePro,$msg)
{

$counPro= mysqli_num_rows($exePro);
if($counPro>0)
{

$classPro=1;
while($resPro = mysqli_fetch_array($exePro))
{
$productcode= $resPro["productcode"];
$auto_number= $resPro["auto_number"];
$productname= $resPro["productname"];
$price= $resPro["price"];
//setlocale(LC_MONETARY,"en_US");
$price = $price;
$show_price= $resPro["show_price"];
$discount = (($show_price-$price)*100)/$show_price;
$img = "defualt.jpg";
$sql = "select * from productsimage where productanum='$auto_number' and size_type='s'";
$exeimg = mysqli_query($con,$sql)or die(mysql_error());



if(mysqli_num_rows($exeimg)>0)
{
$resimg = mysqli_fetch_array($exeimg);
$img = $resimg['imagename'];
}
if($classPro%3==0)
{
$lastclass='no_margin_right';
}else{
$lastclass='';
}

?>

<div class="product_box <?php echo $lastclass;?>" >
<form method="post" action="productdetail.php" id="frmdetail<?php echo $classPro;?>">
<input type="hidden" name="productname" value="<?php echo $productname?>">
<input type="hidden" name="pid" value="<?php echo $productcode?>">
<input type="hidden" name="id" value="<?php echo $auto_number?>">
</form>

<img src="productimages/<?php echo $img?>" alt="Image
<?php echo $classPro;?>" style="height:150px;width:200px;cursor:pointer;" onclick="subform('frmdetail<?php echo $classPro;?>')"; />

<h3><?php echo $productname;?></h3>
<p class="product_price"><span style="text-decoration:line-through;"><?php echo currency.$show_price?></span><br/>
<?php echo currency.$price;?>
</p>

<form method="post" action="shoppingcart.php" id="frmaddproduct<?php echo $classPro;?>">
<input type="hidden" name="product_code" value="<?php echo $productcode?>">
<input type="hidden" name="product_qty" value="1">
<input type="hidden" name="type" value="add">
<input type="hidden" name="pid" value="<?php echo $auto_number;?>">
<input type="hidden" name="return_url" value="<?php echo $current_url;?>">

</form>


<a style="cursor:pointer;" onclick="subform('frmaddproduct<?php echo $classPro;?>')" class="add_to_card">Add to Cart</a>

<a style="cursor:pointer;" onclick="subform('frmdetail<?php echo $classPro;?>')" class="detail">Detail</a>
</div>


<?php
$classPro++;
}
}else
{
//echo "Product is not available";
echo $msg;
}




}

?>

这是截图 enter image description here当我运行应用程序时..

最佳答案

将连接传递给用户定义的函数获取记录

在函数定义中多取一个参数。

function getrecords($exePro,$msg,$con){
//Code.....
}

当您调用 getrecords

getrecodes ($somedata,$somemsg,$con);

当您调用自定义函数时,它会从 connection.php 获取连接并将其作为参数传递给自定义函数,因此您可以在其上使用 $con

关于Php 网页无法找到连接变量,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51228056/

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