简体   繁体   English

我如何在mysql中插入数据并单击一个按钮即可在php中同时检索?

[英]how i can insert data in mysql and retrieve simulteneously in php on the click of one button?

actually i want to send a comment to the each image and it should display just after clicking the button . 实际上,我想向每个图像发送评论,并且应该在单击按钮后立即显示它。 I am able to do insert and retrieve the comment but it require refresh the page and i don't want to refresh....just like orkut.plz help me im new in php... 我能够插入和检索评论,但是它需要刷新页面,而我不想刷新....就像orkut.plz可以帮助我在php中实现新功能一样...

thans to all............ 谢谢大家............

insertimg.php insertimg.php

//________________________________________FOR INSERT COMMENT_____________________________________________________
if (isset($_POST['Submit'])) 
{
$sql = "INSERT INTO comment(imid, comm) values ('".mysql_real_escape_string(stripslashes($_REQUEST['imgId']))."', '".mysql_real_escape_string(stripslashes($_REQUEST['Comment']))."')";
//$sql = "INSERT INTO comment (com) VALUES ($_POST['Comment'])";
//$sql="UPDATE upload SET comm='$_REQUEST['Comment']'WHERE id='$_REQUEST['imgId']'";
if($result = mysql_query($sql ,$conn)) 
  {

     echo "submited:";

   }
else 
   {
    echo "<h1>problem </h1> ".mysql_error();
   }

} 

For display comment.. 用于显示注释。

$page=$_GET["page"];
$sql = "select comm from comment where imid = '".$page."'";

$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not get data: ' . mysql_error());
}
echo"Comments:";
echo "<br>";
echo "<br>";
while($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{

 //echo $row['comm'];
 echo "<textarea name=\"Comment\" style=\"background-color:#81F7BE;\">"; echo $row['comm']; echo "</textarea>";
 //echo "<font>";
 echo "<br>";
 echo "<br>";
}   
mysql_close($conn);

?>

You can solve it with AJAX. 您可以使用AJAX解决它。 You can use something like jQuery Ajax library. 您可以使用类似jQuery Ajax库的工具。

$.ajax({
   type: "POST",
   url: "inserting.php",
   data: "imid=1&comm=Hi",
   success: function(msg){
     alert( "Ajax Response: " + msg );
   }
 });

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM