How do i display images stored as blob data type in mysql with php? [closed]

问题

How do i display an image stored in mySQL database as BLOB ? What it tried so far: 1. Created a new php function/file to get picture (getpicture.php). 2. In the html, I have the following code:

<img src="getpicture.php?id=2" border ="0" height="250" width="250" /> 




/*below is the getpicture.php*/
<?php

@ $db = new MySQLi('localhost','root','','myDatabase');

if(mysqli_connect_errno()) {
    echo 'Connection to database failed:'.mysqli_connect_error();
    exit();
}

if(isset($_GET['People_Id'])) {

    $id = mysqli_real_escape_string($_GET['People_Id']);
    $query = mysqli_query("SELECT * FROM 'people' WHERE 'People_Id' = '$id'");

    while($row = mysqli_fetch_assoc($query)) {
        $imageData =$row['image'];
    }

    header("content-type: image/jpeg");
    echo $imageData;
    echo $id;
}

else {
    echo "Error!";
    echo $id;
}

?>

What's wrong with the codes ? Please help!

回答1:

I answered my own question, it's working now..

Below is the getpicture.php:

<?php

$db = new MySQLi('localhost', '', '', 'mydatabase');

if ($db->connect_errno) {
    echo 'Connection to database failed: '. $db->connect_error;
    exit();
}

if (isset($_GET['id'])) {

$id = $db->real_escape_string($_GET['id']);

$query = "SELECT `Picture` FROM member WHERE `Id` = '$id'";

$result = $db->query($query);

while($row = mysqli_fetch_array($result)) {
    $imageData = $row['Picture'];
    header("Content-type:image/jpeg");
    echo $imageData;
}

} ?>

The php script which retrieve the getpicture.php above looks like this:

echo '<img src="getpicture.php?id=' . htmlspecialchars($_GET["id"]) . '"border ="0" height="250" width="250" />';

Thaank you all for the help

回答2:

This is wrong:

$query = mysqli_query("SELECT * FROM 'people' WHERE 'People_Id' = '$id'");

you use wrong quotes for table name (must be backtick instead of single quote (see tylda ~ key). See docs: http://dev.mysql.com/doc/refman/5.1/en/reserved-words.html

Also this is wrong too

header("content-type: image/jpeg");
echo $imageData;
echo $id;

get rid of last echo $i; and replace it with exit(); otherwise you corrupt the image data stream you just sent.

回答3:

Plenty is wrong.

your SQL is wrong. Remove the single quotes from the table and column and replace with back ticks.
Although it may still work, you should have a couple of new lines after your header
You shouldn't echo out your $id after you echo your image data.
You're checking for the wrong value when you check isset
Also, you should be using OOP for mysqli
Since your image data is only a single row, you don't need to wrap it in a while loop

Here is an updated code example

<?php

$db = new MySQLi('localhost', 'user', 'password', 'myDatabase');

if ($db->connect_errno) {
    echo 'Connection to database failed: '. $db->connect_error;
    exit();
}

if (isset($_GET['id'])) {

    $id = $db->real_escape_string($_GET['id']);
    $result = $db->query("SELECT * FROM `people` WHERE `People_Id` = '$id'");

    $row = $result->fetch_assoc();
    $imageData = $row['image'];

    header("Content-type: image/jpeg\n\n");
    echo $imageData;
}

else {
    echo "Error!";
}

来源：https://stackoverflow.com/questions/25209913/how-do-i-display-images-stored-as-blob-data-type-in-mysql-with-php

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php

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image

blob

jpeg