javascript - How to store webcam captured image path in mysql database -


i've tried following code capture image, stores image in specified folder path of image want store in database doesn't work.the execution of code stops after uploding... message.please me,whats going wrong code.

<!--test.php-->  <?php  session_start();  include 'connection.php';  $name = date('ymdhis');  $newname=mysql_real_escape_string("images/".$name.".jpg");  $file = file_put_contents( $newname, file_get_contents('php://input') );  if (!$file) {  	print "error: failed write data $filename, check permissions\n";  	exit();  }  else  {      $sql="insert entry(images) values('$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";    ?>
<!--index.php-->  <?php  session_start();  if(isset ($_post["send"]))  {      $getname=$_post["myname"];      include 'connection.php';      $idvalue=$_session["myvalue"];      $sql="update entry set name='$getname' id='$idvalue'";      $result=mysql_query($sql)              or die(mysql_error());      if($result)      {          echo "uploaded $_session[myvalue] re ..... ";      }      else      {          echo "$_session[myvalue] nahi hua";      }  }  ?>    <form action="<?php echo $_server["php_self"];?>" method="post">      <input type="text" name="myname" id="myname">      <input type="submit" name="send" id="send">  </form>    <script type="text/javascript" src="webcam.js"></script>  <script language="javascript">  		document.write( webcam.get_html(320, 240) );  </script>  <form>  		<input type=button value="configure..." onclick="webcam.configure()">  		&nbsp;&nbsp;  		<input type=button value="take snapshot" onclick="take_snapshot()">  	</form>    <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 , upload 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 shot  				webcam.reset();  			}  			else {alert("php error: " + msg);  			}  		}  	</script>  <div id="upload_results" style="background-color:#eee;"></div>

instead of giving relative path in file_get_contents() give full path/actual path.

hope might you.


Comments

Post a Comment

Popular posts from this blog

php - failed to open stream: HTTP request failed! HTTP/1.0 400 Bad Request -

this code i'm using find latitude , longitude of place: function coords($address){ echo $url="http://maps.googleapis.com/maps/api/geocode/json?address=".$address; $json = file_get_contents(urlencode($url)); $data = json_decode($json, true); print_r($data['results'][0]['geometry']['location']['lat']); print_r($data['results'][0]['geometry']['location']['lng']); } however returns warning: failed open stream: http request failed! http/1.0 400 bad request if use above code in simple procedure works fine not in function. note: have tried curl_init()... method produced same result? instead of encoding whole url encode address part. following code work fine $add='jamshoro phase 2'; $url="http://maps.googleapis.com/maps/api/geocode/json?address=".urlencode($add); $json = file_get_contents($url); $data = json_decode($json, true); print_r($data['r
Read more

java - How to filter a backspace keyboard input -

i have code gives message when key number pressed: txtvalueofclause.addeventfilter(keyevent.key_typed, new eventhandler<keyevent>() { @override public void handle(keyevent t) { char ar[] = t.getcharacter().tochararray(); char ch = ar[t.getcharacter().tochararray().length - 1]; if (!(ch >= '0' && ch <= '9')) { system.out.println("the char entered not number"); t.consume(); } } }); now if press on wrong button , press backspace remove it, error message. how can add backspace input if statement? i have found answer: txtvalueofclause.addeventfilter(keyevent.key_typed, new eventhandler<keyevent>() { @override public void handle(keyevent t) { char ar[] = t.getcharacter().tochararray(); char ch = ar[t.getcharacter().tochar
Read more

java - Show Soft Keyboard when EditText Appears -

i have code in android app, alert dialog , keyboard, , when dialog shown want keyboard appear. not , not sure why? here code: final alertdialog.builder alert = new alertdialog.builder(this); alert.settitle("title"); alert.setmessage("message"); final edittext input = new edittext(this); input.setinputtype(inputtype.type_class_number); alert.setview(input); input.setonfocuschangelistener(new view.onfocuschangelistener() { @override public void onfocuschange(view v, boolean hasfocus) { if (hasfocus) { log.i(tag, "in focus"); inputmethodmanager inputmethodmanager = (inputmethodmanager) getsystemservice(context.input_method_service); inputmethodmanager.showsoftinput(input, inputmethodmanager.show_implicit); } } }); but keyboard not appear though edit text in focus, , tell appear why, doesn't come up? thanks in advance.
Read more