我正在尝试将产品添加到CMS上的数据库中,我想尝试在另一页上显示产品及其图像。
就将对象添加到数据库而言,一切正常,但是我希望上传的图像以数据库中产品的ID命名,该图像被本地存储在目录中,以后将链接至该目录。记录已被添加到数据库中,但是图像上传器无法正常工作,并且在提交表单时出现以下错误:
我的表布局如下所示:
id l名称l数量l描述l价格
1 l钥匙l 5 l解锁东西l 6
注意:未定义的变量:第17行的L:\ xampp \ htdocs \ 655487 \ addSQL.php中的querynew
注意:未定义的变量:第18行的L:\ xampp \ htdocs \ 655487 \ addSQL.php中的querynew
警告:mysqli :: query():第18行的L:\ xampp \ htdocs \ 655487 \ addSQL.php中的空查询
注意:未定义的变量:第18行的L:\ xampp \ htdocs \ 655487 \ addSQL.php中的querynew
这是通过以下方式提交产品的表格:
<fieldset><legend><span> Add a product to the database </span></legend>
<form id ="productsform" method="post" onsubmit="return false;" enctype="multipart/form-data">
<label> Enter a product name: <input type="text" id="name" name="name"/> </label>
<label> Enter a product quantity: <input type="number" id="quantity" name="quantity"/> </label>
<label> Enter a product description: <input type="text" id="description" name="description"/> </label>
<label> Enter a product price: <input type="text" id="price" name="price"/> </label>
<label> Upload a image of the product: <input type="file" id="file1" name="file"></label>
<input id="submit" name="submit" type="button" class="reg" value="Add Product">
<div id="status"></div>
这是后面的脚本addSQL.php:
<?php
include("dbase/config_database.php");
//Stores all information passed through AJAX into the query
$name = $_POST['name'];
$quantity = $_POST['quantity'];
$description = $_POST['description'];
$price = $_POST['price'];
//Adds information to database
$query = "INSERT INTO products (name, quantity, description, price) VALUES ('$name','$quantity','$description','$price')";
//Runs the query
$result = $mysqli->query($query) OR die("Failed query $query");
echo $mysqli->error."<p>";
$querynew - ("SELECT id as 'collectid' from products WHERE name = '$name'and quantity = '$quantity'and description ='$description'and price = '$price'");
$resultnew = $mysqli->query($querynew) OR die("Failed query $querynew");
while($info = mysqli_fetch_array( $resultnew)){
$productid = $info['collectid'];
}
$image = $_FILES['file1']['name'];
$type = $_FILES['file1']['type'];
$size = $_FILES['file1']['size'];
$tmp_name = $_FILES['file1']['tmp_name'];
$imgpath = "images/".$productid.".jpg";
// Run the move_uploaded_file() function here
$moveResult = move_uploaded_file($tmp_name, $imgpath);
// Evaluate the value returned from the function if needed
$querytwo = ("SELECT * FROM products WHERE name = '$name' and quantity = '$quantity' and description = '$description' and price = '$price'");
$resulttwo = $mysqli ->query($querytwo) OR die ("Failed query $querynew");
$info = array();
while($row = mysqli_fetch_assoc($resulttwo)){
$product = array("id" => $row ['id'],
"name" => $row ['name'],
"quantity" => $row ['quantity'],
"description" => $row ['description'],
"price" => $row ['price'],
);
array_push($info,$product);
}
$json_output = json_encode($info);
echo $json_output;
?>
任何帮助将不胜感激!
您有错字,除了注意到以下内容,我没有测试过您的脚本
$ querynew-(“ SELECT id as'collectid ....
应该
$ querynew =(“” SELECT id as'collectid ...
这就是您的错误,希望对您有所帮助
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