Get image filepath using PHP and display in HTML
Question
I have followed several posts and made progress, but I still am unable to display a jpeg image using the html img src tags which is located in my webroot folder under www/SnapShots.
I would like to use PHP to search a local mysql db and return the newest image filepath. I would then like to use that filepath to display the image.
I have tried using both a separate getImage.php and showimage.html as well as as a single file containing both php and html which I will share below. Currently I am able to echo the correct filepath, which when copied and pasted manually into an img src works, however it does not work by referencing the variable or .php url.
webcam.php:
<?php
echo '<!DOCTYPE html>';
echo '<html>';
echo '<body>';
echo '<h1>Display Recent WebCam Image</h1><br />';
//$id = $_GET['id'];
//Connect to mysql DB
$link = mysql_connect("localhost", "root", "password");
mysql_select_db("temperature_database");
//Select thew newest image filepath
$sql = "SELECT imgFilePath FROM tempLog WHERE datetime=(SELECT MAX(tempLog.datetime) FROM tempLog);";
$result = mysql_query("$sql");
$row = mysql_fetch_assoc($result);
mysql_close($link);
//header("Content-type: image/jpeg");
header("Content-type: text/plain");
$imgpath = $row['imgFilePath'];
//echo $imgpath;
echo '<img src="{$imgpath}", alt="Most recent WebCam Shot", width="800", height="600"><br />';
?>
</body>
</html>
I have tried numerous ways of inserting the $imgpath variable and or referencing it from a separate html page ie: .
A point in the right direction would be much appreciated.
As of now webcam.php outputs the following, so there also appears to be an issue with interpreting the html tags.
<!DOCTYPE html><html><body><h1>Display Recent WebCam Image</h1><br /><img src="{$imgpath}", alt="Most recent WebCam Shot", width="800", height="600"><br /></body>
2 answers
solution1
3 ACCPTED 2016-02-17 18:54:07
Use double quotes instead:
// code
echo "<img src='{$imgpath}' alt='Most recent WebCam Shot' width='800' height='600'><br />";
PHP does not replace variables inside single quoted strings.
Update
If I understand well your question, you want to read the contents of a image and use it in a <img> tag.
You have to split the code in two files (I'll illustrate with a simplification):
index.html
<p>Webcam image:</p>
<img src="view_image.php">
view_image.php
<?php
$image_path = function_to_get_image_path();
header("Content-type: image/jpeg");
readfile($image_path);
This is the approach you could take.
I recommend you start with an easy sample:
Create a folder with three files:
A sample image called image.jpg
A HTML file called index.html with the following code:
<img src="view_image.php">
And a file called view_image.php with this content:
<?php
header("content-type: image/jpeg");
readfile("image.jpg");
Note that php has to be enabled in your web server.
solution2
0 2016-02-18 15:59:18
您也可以像这样使用它：
echo '<img src="'.{$imgpath}.'", alt="Most recent WebCam Shot", width="800", height="600"><br />';
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