How to upload images into MySQL database using PHP

2018-12-31 17:00发布

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问题:

I am trying to save images in my database from HTML form. I have written PHP code to accomplish this task. The program is not generating any error message, but also not inserting image data in MySQL database. Kindly check it. Here i am sharing a excerpt from my code.

        /*-------------------
    IMAGE QUERY 
    ---------------*/


    $file   =$_FILES[\'image\'][\'tmp_name\'];
    if(!isset($file))
    {
      echo \'Please select an Image\';
    }
    else 
    {
       $image_check = getimagesize($_FILES[\'image\'][\'tmp_name\']);
       if($image_check==false)
       {
        echo \'Not a Valid Image\';
       }
       else
       {
        $image = file_get_contents ($_FILES[\'image\'][\'tmp_name\']);
        $image_name = $_FILES[\'image\'][\'name\'];
        if ($image_query = mysql_query (\"insert into product_images values (1,\'$image_name\',$image )\"))
        {
          echo $current_id;
         //echo \'Successfull\';
        }
        else
        {
          echo mysql_error();
        }
       }
   }
        /*-----------------
    IMAGE QUERY END
    ---------------------*/

    <form action=\'insert_product.php\' method=\'POST\' enctype=\'multipart/form-data\' ></br>
            File        : <input type=\'file\' name= \'image\' >
    </form>

Error Message You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near \'\' at line 1

回答1:

Firstly, you should check if your image column is BLOB type!

I don\'t know anything about your SQL table, but if I\'ll try to make my own as an example.

We got fields id (int), image (blob) and image_name (varchar(64)).

So the code should look like this (assume ID is always \'1\' and let\'s use this mysql_query):

$image = addslashes(file_get_contents($_FILES[\'image\'][\'tmp_name\'])); //SQL Injection defence!
$image_name = addslashes($_FILES[\'image\'][\'name\']);
$sql = \"INSERT INTO `product_images` (`id`, `image`, `image_name`) VALUES (\'1\', \'{$image}\', \'{$image_name}\')\";
if (!mysql_query($sql)) { // Error handling
    echo \"Something went wrong! :(\"; 
}

You are doing it wrong in many ways. Don\'t use mysql functions - they are deprecated! Use PDO or MySQLi. You should also think about storing files locations on disk. Using MySQL for storing images is thought to be Bad Idea™. Handling SQL table with big data like images can be problematic.

Also your HTML form is out of standards. It should look like this:

<form action=\"insert_product.php\" method=\"POST\" enctype=\"multipart/form-data\">
    <label>File: </label><input type=\"file\" name=\"image\" />
    <input type=\"submit\" />
</form>

Sidenote:

When dealing with files and storing them as a BLOB, the data must be escaped using mysql_real_escape_string(), otherwise it will result in a syntax error.



回答2:

Just few more details：

  • Add mysql field

`image` blob

  • Get data from image

$image = addslashes(file_get_contents($_FILES[\'image\'][\'tmp_name\']));

  • Insert image data into db

$sql = \"INSERT INTO `product_images` (`id`, `image`) VALUES (\'1\', \'{$image}\')\";

  • Show image to the web

<img src=\"data:image/png;base64,\'.base64_encode($row[\'image\']).\'\">

  • End


回答3:

This is the perfect code for uploading and displaying image through MySQL database.

<html>
<body>
<form method=\"post\" enctype=\"multipart/form-data\">
<input type=\"file\" name=\"image\"/>
<input type=\"submit\" name=\"submit\" value=\"Upload\"/>
</form>
<?php
    if(isset($_POST[\'submit\']))
    {
     if(getimagesize($_FILES[\'image\'][\'tmp_name\'])==FALSE)
     {
        echo \" error \";
     }
     else
     {
        $image = $_FILES[\'image\'][\'tmp_name\'];
        $image = addslashes(file_get_contents($image));
        saveimage($image);
     }
    }
    function saveimage($image)
    {
        $dbcon=mysqli_connect(\'localhost\',\'root\',\'\',\'dbname\');
        $qry=\"insert into tablename (name) values (\'$image\')\";
        $result=mysqli_query($dbcon,$qry);
        if($result)
        {
            echo \" <br/>Image uploaded.\";
            header(\'location:urlofpage.php\');
        }
        else
        {
            echo \" error \";
        }
    }
?>
</body>
</html>


回答4:

How to INSERT INTO DB?

<html>
    <head>
        <title>Uploads</title>
    </head>
    <body>
        <input type=\"file\" name=\"file_array[]\"/>
        <input type=\"file\" name=\"file_array[]\"/>
        <input type=\"file\" name=\"file_array[]\"/>
        <input type=\"submit\" name=\"submit\"/>
    </body>
</html>

<?php
    if(isset($_FILES[\'file_array\']))
    {
        $name_array = $_FILES[\'file_array\'][\'name\'];
        $tmp_name_array = $_FILES[\'file_array\'][\'tmp_name\'];
        $type_array = $_FILES[\'file_array\'][\'type\'];
      $size_array = $_FILES[\'file_array\'][\'size\'];
        $error_array = $_FILES[\'file_array\'][\'error\'];
        $dir = \"slideshow\";

        for($i = 0; $i<count($tmp_name_array); $i++)
        {
            if(move_uploaded_file($tmp_name_array,\"slideshow/\".$name_array))
            {
                echo $name_array[$i].\"Upload is complete<br>\";

            }  else {
                echo\"Move_uploaded_file function failed for\".$name_array[$i].\"<br>\";
            }
        }
    }
?>


标签： php html mysql database image-uploading
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