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我想将图像上传到我的数据库,所以为此我创建了如下表格。图像(BLOB)不能完美地将图像保存在数据库中
而且使用下面的PHP代码上传并显示在同一页上的图像。
<?php
// Connection To Database
$host = "localhost";
$username = "root";
$password = "";
$dbname = "my_db";
$connection = mysqli_connect($host,$username,$password,$dbname);
if (!$connection)
{
die('Could Not Connect To Database: ' . mysqli_connect_error());
}
mysqli_query($connection, "SET NAMES utf8");
// Garb User Account Details
$garb = mysqli_query($connection, "SELECT * FROM my_table");
if(!$garb){
$error = "<div class='error'>There's Little Problem: ".mysql_error()."</div>";
} else {
//Showing The User Data
while($row = mysqli_fetch_array($garb)) {
$A_Id = $row['A_Id'];
$A_ProfilePic = $row['A_ProfilePic'];
}
// Getting Image
$A_ProfilePic = addslashes(file_get_contents($_FILES['A_ProfilePic']['tmp_name'])); //SQL Injection defence!
$A_ProfilePic = mysqli_real_escape_string($connection, $A_ProfilePic);
//Everything Is Okay So Let's Register This User
$update = mysqli_query($connection, "UPDATE accounts SET A_ProfilePic='$A_ProfilePic' WHERE A_Id='999999'");
if(!$update){
$updateerror = "<div class='error'>There's Little Problem: ".mysql_error()."</div>";
} else {
//Confirming Message To User
$error = "<div class='success'>Your Pic Is Updated.</div><br/>";
}
?>
/* Getting Image */
<?php echo '<img class="profileAvatar" alt="" title="" src="data:image/jpeg;base64,'.base64_encode($A_ProfilePic).'"/>';?>
<form action="" method="post" enctype="multipart/form-data">
<input type="file" name="A_ProfilePic" id="A_ProfilePic"/>
<input type="submit" name="submit" value="Update Your Setting"></input>
<input type="reset" name="reset" value="Reset Form"></input>
</form>
但我得到的错误作为插入图像,然后,当我的Base64编码不完全插入,尽管检索图像为什么我不能够看到它。我从在线Base64转换器转换了相同的图像,然后从我的数据库代码中获得了不同的代码。那么最新的错误...?
Your'e用'addslashes()'和'mysqli_real_escape_string()'双重转义。只要使用准备好的声明并停止这样做。 – AbraCadaver
我也尝试了一个,但没有工作... –
插入的那些已经与双重逃脱搞砸了,如果你在显示它之前逃脱它将不会被纠正。 – AbraCadaver