2016-12-31 34 views
0

我一直在環顧四周的論壇,並嘗試了一切,但我不能讓我的按鈕禁用後點擊它,我剛開始使用PHP,這是我的第一個項目之一。它是一個喜歡和不喜歡的系統。有人知道這個遊戲機嗎? 當您按下類似按鈕或不喜歡按鈕時,我希望禁用這兩個按鈕。1單擊然後禁用 - PHP

<!--Like System--> 
<center> 

<?php 

$sql = "SELECT * FROM likes"; 
$result = mysqli_query($con, $sql); 

if (mysqli_num_rows($result) > 0) { 
    // output data of each row 
    while($row = mysqli_fetch_assoc($result)) { 
     ?> <h1 class="lcount"><?php echo $row["average"];?> <span class="glyphicon glyphicon-signal"></span></h1></br><?php 

    } 
} else { 
    echo "0 results"; 
} 
?> 

<?php 

$sql = "SELECT * FROM likes"; 
$result = mysqli_query($con, $sql); 

if (mysqli_num_rows($result) > 0) { 
    // output data of each row 
    while($row = mysqli_fetch_assoc($result)) { 
     ?> <h1 class="lcount"><?php echo $row["like"];?> 
     <span style="color: green;"class="glyphicon glyphicon-thumbs-up"></span></h1></br> 

<?php 
    } 
} else { 
    echo "0 results"; 
} 
?> 

<?php 

$sql = "SELECT * FROM likes"; 
$result = mysqli_query($con, $sql); 

if (mysqli_num_rows($result) > 0) { 
    // output data of each row 
    while($row = mysqli_fetch_assoc($result)) { 
     ?> <h1 class="lcount"><?php echo $row["dislike"];?> 
     <span style="color: red;" class="glyphicon glyphicon-thumbs-down"></span></h1> 

<?php 
    } 
} else { 
    echo "0 results"; 
} 
?> 

<?php 

if(isset($_POST['like'])){ 

    $sql = "UPDATE `likes` SET `like` = `like` + 1 , `average` = `average` + 1"; 


    if (mysqli_query($con, $sql)) { 
    } else { 
     echo "Error updating record: " . mysqli_error($con); 
    } 
} 
?> 

<?php 

if(isset($_POST['dislike'])){ 

    $sql = "UPDATE `likes` SET `dislike` = `dislike` + 1 , `average` = `average` - 1"; 


    if (mysqli_query($con, $sql)) { 
    } else { 
     echo "Error updating record: " . mysqli_error($con); 
    } 
} 
?> 

    <form type="#" method="post"> 
     <button name="like" type="submit" class="btn btn-success"> 
      <span class="glyphicon glyphicon-thumbs-up"></span></button> 

     <button name="dislike" type="submit" class="btn btn-danger"> 
      <span class="glyphicon glyphicon-thumbs-down"></span></button> 
    </form> 
</center> 
<!--Like System END--> 
+3

我覺得像JavaScript將這個 – Maciej

+0

着它只有PHP的工作更適合? – FluffyMe

+2

PHP需要您重新加載頁面,因爲代碼在服務器上執行。 Javascript在瀏覽器中執行,因此它更適合操作UI,例如禁用按鈕。理論上,你可以在PHP中做到這一點,但它可能會非常混亂,總體上不值得。 – Maciej

回答

0

如果你只想PHP使用像這樣:

<?php 

$status = 'enabled'; 

if (isset($_POST['like'])) { 
    $status = 'disabled'; 
} 


?> 

<form type="#" method="post"> 
    <button name="like" type="submit" class="btn btn-success" <?=$status?>> 
       Button is <?=$status?> 
    </button> 
</form> 

您可以使用jQuery像這樣:

https://jsfiddle.net/y4qLew3s/1/

更新時間:

我更新了。它的工作原理與喜歡或不喜歡按鈕:

https://jsfiddle.net/y4qLew3s/3/

+0

以及這工作,但只有被點擊的按鈕將被禁用是否有一種方法來禁用不喜歡按鈕? – FluffyMe

+0

@FluffyMe我更新了代碼:https://jsfiddle.net/y4qLew3s/3/ – Ali

+0

按鈕的類必須保持他們的方式,因爲即時通訊使用引導 – FluffyMe

0

在JavaScript

setAttribute()

function disable(){ 
    document.getElementById('submit').setAttribute("disabled", "disabled"); 

}

更新 如果你在同一時間

禁用這兩個按鈕
<script> 
function disable(){ 
    document.getElementById('submit').setAttribute("disabled", "disabled"); 
    document.getElementById('submit1').setAttribute("disabled", "disabled"); 
} 
</script> 

<input title="" id="submit" type="submit" name="button" onclick="disable()" value="like"> 

<input title="" id="submit1" type="submit" name="button" onclick="disable()" value="unlike"> 

check here

+0

'submit'不是編號 –

+0

@KOUROUMA就是這樣的例子。 – denny

+0

仍然沒有工作 – FluffyMe