-5

I am using the following code to change button when it is clicked by echoing the javascript's document.getElemenById property but the button remains as it is:

while($result=mysql_fetch_array($result1))
                    {
                        $i=1;
                        echo "
                        <tr>
                        <form action='student.php' method='post'>
                        <td>$result[0]<input type=\"hidden\" name=\"company\" value=\"$result[0]\"></td>
                        <td>
                        <input type=\"checkbox\" name=\"yes\" />
                        <input type=\"submit\" name=\"apply$i\" id=\"apply$i\" class=\"buttons\" style=\"border: none\" value=\"Apply now!\" 
                                  onclick=\"return confirm('Are You Sure you want to apply?');\">   
                        <input type=\"hidden\" name=\"apply$i\" value=\"apply$i\">
                        </td>
                        </form>
                        </tr>
                        <tr><td><input type=\"text\" id=\"ankur\"></td></tr>
                        ";
                        $i++;
                    }
                    echo "</table></center>";

Here the button with the id apply$i should change to Applied string as per the following code

$id=$_SESSION['sessionid'];
if(isset($_POST['apply1']))
{
    if(isset($_POST['yes']))
    {
        $yes=trim(mysql_prep($_POST['yes']));
        $company=trim(mysql_prep($_POST['company']));
        $button_id=trim(mysql_prep($_POST['apply1']));
        if($yes=="on")
        {
            $query ="select `Branch`,`Class` from `student_details` where `Userid`='$id'"; 
            $temp1= mysql_query($query);
            $temp2= mysql_fetch_array($temp1);
            $query="insert into `student_applied` values ('$id' ,'$temp2[0]' ,'$temp2[1]' ,'$company')";
            //$query="UPDATE `student`.`users` SET  `Activated` =  '1' WHERE  `users`.`Userid` =  '$student'";
            $result = mysql_query($query);
            confirm_query($result);
            if($result)
            {
            echo "ok";
            echo "<script language='javascript' type='text/javascript'>document.getElementById('apply1').innerHTML ='Applied';</script>"

Does anybody know the reason?Plz help

4

1 回答 1

1

而不是在循环内设置g $i，你应该在外面设置它然后增加值，比如

$i=1;
while($result=mysql_fetch_array($result1)){
    //Your rest of code to output content
    $i++;
}

这样，您可以为每个按钮设置不同的 ID，并且 Javascript 应该非常适合您。

于 2013-05-02T18:27:30.807 回答