Question

我为我的梦幻足球网站创建了一个团队建设者。它适用于Chrome和Firefox，但不适用于IE。当我按下IE上的其中一个提交按钮时，整个页面将重新加载该帧，而不是该帧中的内容。

测试网站的链接：http://fantasysoccerpro.com/project/test-ny/

IE bug的屏幕截图：http://fantasysoccerpro.com/iebug.jpg

//SELECT GOALKEEPER//

$database = "wp_fsprowpdatatable_3";
$goalkeeper="SELECT * FROM `fantasys_fspro`. `$database` where POSITION =     'G'";
$gk = mysqli_query(  $con, $goalkeeper );
?>

  <!-- Example row of columns -->

    <form method="post" action=" ">


        <select name="gk_">
        <option>----------Please Select G----------</option>
  <?php while($gkrow = mysqli_fetch_array($gk))
{ ?>
<option value="<?php echo $gkrow['wdt_ID'];?>"> <?php echo    $gkrow['player'];?></option>
<?php }?>              
        </select>
      </div>


      <button type="submit" class="btn btn-info" name="register" >Submit</button>


    </form>
<?php
if(isset($_POST["register"]))
{

$gk_=$_POST["gk_"];

if ($gk_ > "0") {$selectedgk=$gk_;
}}

//UPDATE TABLE WITH GOALKEEPER//


$sql = "UPDATE $pw 
SET PLAYER= (SELECT PLAYER 
         FROM $database
         WHERE wdt_ID=$gk_) WHERE id = 1";

if ($gk_ > "0") {

if ($con->query($sql) === TRUE) {

} else {
echo "Error updating record: " . $con->error;
}}

$sql = "UPDATE $pw
SET SALARY= (SELECT SALARY 
         FROM $database
         WHERE wdt_ID=$gk_) WHERE id = 1";

if ($gk_ > "0") {

if ($con->query($sql) === TRUE) {

} else {
echo "Error updating record: " . $con->error;
}}

$sql = "UPDATE $pw 
SET PROJECTION= (SELECT PROJECTION
         FROM $database
         WHERE wdt_ID=$gk_) WHERE id = 1";

if ($gk_ > "0") {

if ($con->query($sql) === TRUE) {

} else {
echo "Error updating record: " . $con->error;
}}

Answer 1

在iframe中，您有一个包含单个空格的action属性的表单。 Internet Explorer将此解释为应该发布到页面根目录而不是iframe源的表单。完全删除action属性，如@chiliNUT所示，以解决您的问题。

提交按钮的IE错误

1 个答案: