Question

我知道将图像存储在mysql数据库中并不是一个好主意，但我只想尝试一下。

在mysql中我创建了这个表：

CREATE TABLE tbl_images (
      id tinyint(3) unsigned NOT NULL auto_increment,
      image mediumblob NOT NULL,
      PRIMARY KEY (id)
      );

这是php代码：

 if(isset($_FILES['image']) && $_FILES['image']['size'] > 0) {
$tmpName  = $_FILES['image']['tmp_name'];

      $fp   = fopen($tmpName, 'r');
      $data = fread($fp, filesize($tmpName));
      $data = addslashes($data);
      fclose($fp);

      $query ="INSERT INTO tbl_images(image)VALUES('".$data."')";
      $results=mysql_query($query) or die(mysql_error());
       $num=mysql_num_rows($results);
       if($num>0)
           print "Thank you, your file has been uploaded.";
}
else {
   print "No image selected/uploaded";
}
?>

我在mysta的sytax中有错误，这里是输出：

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in ...\insertimg.php on line 21

<html>
<title>Image</title>
<!--mikhaim ie form me3 upload ghabli benevisim-->
<!-- inbar mikhaim 2 database image ra gharar dahim-->
<form enctype="multipart/form-data" action="insertimg.php" method="post" name="changer">
    <input name="MAX_FILE_SIZE" value="102400" type="hidden">
    <input name="image" id="image" accept="image/jpeg" type="file">
    <input value="Submit" type="submit">
    </form>
    </html>

Answer 1

你的Else支架没有关闭.....

Answer 2

在INSERT查询上检查num_rows没有意义。如果您想在错误发生时打印错误，请执行：mysql_query(...) or die(mysql_error())。

编辑：我真的不明白你的INSERT查询为什么会失败，所以你可以删除所有的错误检查，然后在查询后打印“谢谢”。

编辑：这是它应该是什么样子：

<?php
if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) {
    $tmpName = $_FILES['image']['tmp_name'];
    $fp = fopen($tmpName, 'r');
    $data = fread($fp, filesize($tmpName));
    $data = addslashes($data);
    fclose($fp);

    $query = "INSERT INTO tbl_images (image) VALUES ('$data')";
    mysql_query($query) or die(mysql_error());
    print "Thank you, your file has been uploaded.";
else {
    print "No image selected/uploaded";
}

Answer 3

确保记录正在插入，如果插入正常，则可以使用

$num = msql_affected_rows();

或

$num = mysql_insert_id();

而不是$num=mysql_num_rows($results);

语法mysql / php错误图像

3 个答案: