Question

我想从网络摄像头用户图像中捕获图像，使用php将存储在指定文件夹中的图像和捕获的图像路径存储到mysql中。我有一个问题，网络摄像头捕获的图像路径没有存储在mysql数据库中。所以请帮帮我......

＆＃13;

 
    <script type="text/javascript" src="webcam.js"></script>
    <script language="JavaScript">
    		document.write( webcam.get_html(320, 240) );
    </script>
<script language="JavaScript">
        webcam.set_api_url( 'test.php' );
    		webcam.set_quality( 90 ); // JPEG quality (1 - 100)
    		webcam.set_shutter_sound( true ); // play shutter click sound
    		webcam.set_hook( 'onComplete', 'my_completion_handler' );

    		function take_snapshot(){
    			// take snapshot and upload to server
    			document.getElementById('upload_results').innerHTML = '<h1>Uploading...</h1>';
    			webcam.snap();
    		}

    		function my_completion_handler(msg) {
    			// extract URL out of PHP output
    			if (msg.match(/(http\:\/\/\S+)/)) {
    				// show JPEG image in page
    				document.getElementById('upload_results').innerHTML ='<h1>Upload Successful!</h1>';
    				// reset camera for another shot
    				webcam.reset();
    			}
    			else {alert("PHP Error: " + msg);
    			}
    		}
    	</script>

＆＃13;

-------------------------------------
____________test.php_________________
    <?php
session_start();
include 'connection.php';
$name = date('YmdHis');
$newname="images/".$name.".jpg";
$file = file_put_contents( $newname, file_get_contents('php://input') );
if (!$file) {
	print "ERROR: Failed to write data to $filename, check permissions\n";
	exit();
}
else
{
    $sql="Insert into entry(images) values('$newname')";
    $result=mysql_query($con,$sql)
            or die("Error in query");
    $value=mysql_insert_id($con);
    $_SESSION["myvalue"]=$value;
}

$url = 'http://' . $_SERVER['HTTP_HOST'] . dirname($_SERVER['REQUEST_URI']) . '/' . $newname;
print "$url\n";

?>

＆＃13;

 
   <form>
   <input type=button value="Configure..." onClick="webcam.configure()">
    		&nbsp;&nbsp;
   <input type=button value="Take Snapshot" onClick="take_snapshot()">
   </form>
    <div id="upload_results" style="background-color:#eee;"></div>

＆＃13;

Answer 1

所以你说一切都有效，除了＆＃34;插入条目（图像）值（＆＃39; $ newname＆＃39;）＆＃34;？

尝试回显完整的mySQL错误而不是简短的＆＃34;查询错误＆＃34;。它可能是表不存在，列不存在，它没有自动增量列，另一个未命名列没有默认值......可能是任何东西。

如何使用php将scam捕获的图像存储在指定文件夹中并在mysql数据库中捕获图像路径..？

1 个答案: