【发布时间】：2019-07-10 19:12:34
【问题描述】：
当我运行代码时，它在第 135 行给出了类 mysqli_connect 对象的错误无法转换为字符串。请有人为我提供此问题的解决方案。
我尝试寻找解决方案，但没有人为我工作。下面是显示错误的代码。
$row = mysqli_connect("$con,$insert_product");
接受运行但在第 135 行显示错误。我犯了什么错误。完整代码在这里
<?php require_once ("includes/db.php");
?>
<!DOCTYPE html>
<html>
<head>
<title> product</title>
</head>
<body>
<form method="post" action="insert_product.php" enctype="multipart/form-data">
<table width="700" align="center">
<tr>
<td><h2>insert new product</h2></td>
</tr>
<tr>
<td>Product title</td>
<td><input type="text" name="product_title"></td>
</tr>
<tr>
<td>Product category</td>
<td><select name="product_cat">
<option>Select a category</option>
<?php
$get_cats = "select * from category";
$run_cats= mysqli_query($con,$get_cats);
while ($row_cats=mysqli_fetch_array($run_cats))
{
$cat_id = $row_cats['cat_id'];
$cat_title=$row_cats['cat_title'];
echo "<option value='$cat_id'>$cat_title</option>";
}
?>
</select>
</td>
</tr>
<tr>
<td>Product brand</td>
<td><select name="product_brand">
<option>Select brand</option>
<?php
$get_brand = "select * from brand";
$run_brand= mysqli_query($con,$get_cats);
while ($row_brand=mysqli_fetch_array($run_brand))
{
$brand_id = $row_brand['brand_id'];
$brand_title=$row_brand['brand_title'];
echo "<option value='$cat_id'>$cat_title</option>";
}
?>
</select>
</td>
</tr>
<tr>
<td>product image1</td>
<td><input type="file" name="product_img1"></td>
</tr>
<tr>
<td>Product image2</td>
<td><input type="file" name="product_img2"></td>
</tr>
<tr>
<td>Product image3</td>
<td><input type="file" name="product_img3"></td>
</tr>
<
<tr>
<td>Product price</td>
<td><input type="number" name="product_price"></td>
</tr>
<tr>
<td>Product desceptration</td>
<td><input type="text" name="product_desc"></td>
</tr>
<tr>
<td>Product keyword</td>
<td><input type="number" name="product_keyword"></td>
</tr>
<tr>
<td><input type="submit" value="submit" name="submit"></td>
</tr>
</table>
</form>
</body>
</html>
<?php
if(isset($_POST['submit']))
{
$product_title=$_POST['product_title'];
$product_cat=$_POST['product_cat'];
$product_brand=$_POST['product_brand'];
$product_price=$_POST['product_price'];
$product_desc=$_POST['product_desc'];
$status='on';
$product_keyword=$_POST['product_keyword'];
//image name
$product_img1 = $_FILES['product_img1']['name'];
$product_img2 = $_FILES['product_img2']['name'];
$product_img3 = $_FILES['product_img3']['name'];
//image temp names
$temp_name1 = $_FILES['product_img1']['tmp_name'];
$temp_name2 = $_FILES['product_img2']['tmp_name'];
$temp_name3 = $_FILES['product_img3']['tmp_name'];
if($product_title=='' OR $product_cat=='' OR $product_brand=='' OR $product_price=='' OR $product_desc=='' OR $product_keyword=='' OR $product_img1=='' OR $product_img2=='' OR $product_img3=='')
{
echo "<script> alert ('please insert the data in the form')</script>";
exit();
}
else{
//uploadinimage to the folder
move_uploaded_file($temp_name1,"product_images/$product_img1");
move_uploaded_file($temp_name2, "product_images/$product_img2");
move_uploaded_file($temp_name3,"product_images/$product_img3");
}
$insert_product = "insert into products (cat_id,brand_id,date,product_title,product_img1,product_img2,product_img3,product_brand,product_price,product_desc,status) values ('$product_cat','$product_brand',NOW(),'$product_title','$product_img1','$product_img2',$product_img3,'product_brand','product_price','product_desc','product_keyword')";
$row = mysqli_connect("$con,$insert_product");
if($row)
{
echo "<script> alert('insert sucessfully')</script>";
}
else
{
echo "<script> alert(' unsucessfull to insert')</script>";
}
}
?>
【问题讨论】：
-
其他查询是否有效？我看不到连接器的代码，我假设它包含在 db.php 中。
-
你在 $row = mysqli_connect("$con,$insert_product"); 中有引号试试 $row = mysqli_query($con, $insert_product);
-
而不是 $row = mysqli_connect("$con,$insert_product");使用 $row = mysqli_query($con, $insert_product);。您无需再次连接。只需要执行查询。还要删除使这些变量成为字符串的引号。
标签： php string class object mysqli