我已经尝试了以下代码来捕获图像,它将图像存储在指定的文件夹中,但我想要存储在数据库中的图像路径不起作用。代码的执行在Uploding之后停止。 .. message.Please帮我,我的代码出了什么问题。
<!--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 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";
?>
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<!--index.php-->
<?php
session_start();
if(isset ($_POST["send"]))
{
$getname=$_POST["myname"];
include 'connection.php';
$idvalue=$_SESSION["myvalue"];
$sql="update entry set name='$getname' where 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()">
<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 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>
<div id="upload_results" style="background-color:#eee;"></div>
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答案 0 :(得分:1)
而不是在file_get_contents()
给出相对路径给出完整路径/实际路径。
希望这对你有所帮助。