【问题标题】:How to insert data into table in mysqli?如何将数据插入到mysqli中的表中?
【发布时间】:2019-06-02 19:32:35
【问题描述】:

我可以看到我在表单中输入的所有数据,但无法将其插入到表格中。

<?php
echo "u r at starting";
$product_title = $_POST["product_title"];
$product_cat = $_POST["product_cat"];
$cat = $_POST["cat"];
$product_price = $_POST["product_price"];
$product_keywords = $_POST["product_keywords"];
$product_desc = $_POST["product_desc"];

$product_img1 = $_FILES["product_img1"]["name"];
$product_img2 = $_FILES["product_img2"]["name"];
$product_img3 = $_FILES["product_img3"]["name"];

echo $product_title;
echo $product_cat;
echo $cat;
echo $product_price;
echo $product_keywords;
echo $product_desc;
echo $product_img1;
echo $product_img2;
echo $product_img3;


if(isset($_POST["abc"]))
{

$temp_name1 = $_FILES["product_img1"]["tmp_name"];
$temp_name2 = $_FILES["product_img2"]["tmp_name"];
$temp_name3 = $_FILES["product_img3"]["tmp_name"];

move_uploaded_file($temp_name1, "product_images/img1");
move_uploaded_file($temp_name2, "product_images/img2");
move_uploaded_file($temp_name3, "product_images/img3");

$insert_product = "INSERT INTO product_tab VALUES('$product_cat','$cat_id',NOW(),'$product_title','$product_img1','$product_img2','$product_img3','$product_price','$product_keywords','$product_desc')";

$run_product = mysqli_query($con,$insert_product);
echo $run_product;
if($run_product)
{

    echo "Product inserted successfully";
}
else
{
    echo "NOt inserted";
}
}
echo "end";
?>

【问题讨论】:

标签: php sql mysqli


【解决方案1】:

$con 是脚本中定义的变量吗?您显示的代码没有定义 $con。

【讨论】:

  • 是的,我在文件顶部包含了数据库连接代码
猜你喜欢
  • 1970-01-01
  • 2019-03-17
  • 1970-01-01
  • 2013-05-26
  • 1970-01-01
  • 1970-01-01
  • 2014-07-26
相关资源
最近更新 更多