如何使按鈕在同一頁面上執行，但不在下一頁

用戶點擊提交按鈕後，我有一個頁面，然後按鈕將表單內容提交到下一頁執行並保存到數據庫。

因此，我的頁面將轉到另一頁執行並返回到目標頁面。

如：index.php文件和exec.php

的index.php：

<form name="g-form" action="gbtn-exec.php" method="post" class="goat-vote" onsubmit="return validategForm()"> 
<input type="text" name="g-product" placeholder="Brand/Product Name" style="-moz-border-radius: 5px; border-radius: 5px; padding-left:20px; opacity:.5; border:none; margin-left:110px; width:440px; height:38px; font-family:'Proxima Nova Rg';color:#000; font:1.6em;" /> 


<p class="g-question">Why you love it?</p> 

<textarea name="g-reason" style="-moz-border-radius: 5px; border-radius: 5px; padding:5px; opacity:.5; border:none; margin-left:110px; width:450px; height:150px; font-family:'Proxima Nova Rg';color:#333; font-size:1em;"></textarea> 

<input name="g-btn" class="vote-btn" type="submit" value="vote" style="margin-left:470px; cursor:pointer;"></form>

exec.php

if ($_POST["g-product"] && $_POST["g-reason"] != "") 
{ 
$gproduct = $_POST["g-product"]; 
$greason = $_POST["g-reason"]; 

$insert ="INSERT INTO jovine.vote (vote_id ,product_name ,reason ,type) VALUES (NULL, '$gproduct', '$greason', 'goat')"; 
$result = mysql_query($insert,$con); 
echo "<script>"; 
echo "alert('Thank you. Your vote has been recorded.');"; 
echo "window.location.href='index.php';"; 
echo "</script>"; 
}

我的問題是，我怎麼能執行的提交按鈕沒有去exec.php的index.php？（意味着將兩者結合在一個.php中）這是因爲當用戶點擊提交按鈕時，它會進入空白頁面執行，看起來不太好。

任何人都可以提供幫助嗎？謝謝！ =）

來源

2013-09-23 youaremysunshine

使用ajax方法在後臺執行exec.php並在index.php中檢索結果 –

嘗試使用jquery ajax – iJade

@DonovanCharpin我可以知道該如何編寫腳本？我有點不明白maunal ... =（ – youaremysunshine

if (isset($_POST["g-btn"])) 
{ 
$gproduct = $_POST["g-product"]; 
$greason = $_POST["g-reason"]; 

$insert ="INSERT INTO jovine.vote (vote_id ,product_name ,reason ,type) VALUES (NULL, '$gproduct', '$greason', 'goat')"; 
$result = mysql_query($insert,$con); 
echo "<script>"; 
echo "alert('Thank you. Your vote has been recorded.');"; 
echo "window.location.href='index.php';"; 
echo "</script>"; 
} 

<html> 
<form name="g-form" action="a.php" method="post" class="goat-vote" onsubmit="return validategForm()"> 
<input type="text" name="g-product" placeholder="Brand/Product Name" style="-moz-border-radius: 5px; border-radius: 5px; padding-left:20px; opacity:.5; border:none; margin-left:110px; width:440px; height:38px; font-family:'Proxima Nova Rg';color:#000; font:1.6em;" /> 


<p class="g-question">Why you love it?</p> 

<textarea name="g-reason" style="-moz-border-radius: 5px; border-radius: 5px; padding:5px; opacity:.5; border:none; margin-left:110px; width:450px; height:150px; font-family:'Proxima Nova Rg';color:#333; font-size:1em;"></textarea> 

<input name="g-btn" class="vote-btn" type="submit" value="vote" style="margin-left:470px; cursor:pointer;"></form> 
</html>

你可以這樣做... isset（$ _ POST [「g-btn」]）會檢查它是否被點擊或不是？

來源

2013-09-23 08:01:02 Kalpit

編輯：忽略此請，沒有看過你的問題完全，因爲它似乎

嘗試是這樣的：

的index.php

<form name="g-form" action="gbtn-exec.php" method="post" class="goat-vote" onsubmit="return validategForm()"> 
    <input type="text" name="g-product" placeholder="Brand/Product Name" style="-moz-border-radius: 5px; border-radius: 5px; padding-left:20px; opacity:.5; border:none; margin-left:110px; width:440px; height:38px; font-family:'Proxima Nova Rg';color:#000; font:1.6em;" /> 

    <p class="g-question">Why you love it?</p> 

    <textarea name="g-reason" style="-moz-border-radius: 5px; border-radius: 5px; padding:5px; opacity:.5; border:none; margin-left:110px; width:450px; height:150px; font-family:'Proxima Nova Rg';color:#333; font-size:1em;"></textarea> 

    <?php if(!isset($_GET['hidesubmit'])): ?> 
    <input name="g-btn" class="vote-btn" type="submit" value="vote" style="margin-left:470px; cursor:pointer;"> 
    <?php endif; ?> 
</form>

高管。 php

if ($_POST["g-product"] && $_POST["g-reason"] != "") 
{ 
    $gproduct = $_POST["g-product"]; 
    $greason = $_POST["g-reason"]; 

    $insert ="INSERT INTO jovine.vote (vote_id ,product_name ,reason ,type) VALUES (NULL, '$gproduct', '$greason', 'goat')"; 
    $result = mysql_query($insert,$con); 
    echo "<script>"; 
    echo "alert('Thank you. Your vote has been recorded.');"; 
    echo "window.location.href='index.php?hidesubmit=1';"; 
    echo "</script>"; 
}

來源

2013-09-23 08:00:40 Joshua

對不起，但爲什麼它不工作？ – youaremysunshine

它應該工作，但你想要一個解決方案，無需發送表單 - 我的解決方案發送表單並重定向到原始頁面特殊參數，在此參數的基礎上顯示提交按鈕。 – Joshua

添加的jQuery的lib在你的頭和你的jQuery perso（myjQuery.js），您將使用後的ajax

的index.php

<script src="http://code.jquery.com/jquery-1.10.1.min.js"></script> 
<script src="/js/myjQuery.js"></script>

沒有你的表格上的操作創建表單。

<html> 
    <form name="g-form" class="goat-vote"> 
     <input type="text" name="g-product" placeholder="Brand/Product Name" style="-moz-border-radius: 5px; border-radius: 5px; padding-left:20px; opacity:.5; border:none; margin-left:110px; width:440px; height:38px; font-family:'Proxima Nova Rg';color:#000; font:1.6em;" /> 
     <p class="g-question">Why you love it?</p> 
     <textarea name="g-reason" style="-moz-border-radius: 5px; border-radius: 5px; padding:5px; opacity:.5; border:none; margin-left:110px; width:450px; height:150px; font-family:'Proxima Nova Rg';color:#333; font-size:1em;"></textarea> 
     <input name="g-btn" class="vote-btn" value="vote" style="margin-left:470px; cursor:pointer;"> 
    </form> 
</html>

讓您exec.php這樣

exec.php

if ($_POST["g-product"] && $_POST["g-reason"] != "") 
{ 
    $gproduct = $_POST["g-product"]; 
    $greason = $_POST["g-reason"]; 
    $insert ="INSERT INTO jovine.vote (vote_id ,product_name ,reason ,type) VALUES (NULL, '$gproduct', '$greason', 'goat')"; 
    $result = mysql_query($insert,$con); 
    echo "<script>"; 
    echo "alert('Thank you. Your vote has been recorded.');"; 
    echo "window.location.href='index.php?hidesubmit=1';"; 
    echo "</script>"; 
}

，並創建myjQuery.js

myjQuery.js

$(document).ready(function(){ 
    $(".vote-btn").on("click", function(){ 
     g-product = $("input[name='g-product']").val(); 
     g-reason = $("input[name='g-reason']").val(); 
     $.ajax({cache:false,dataType:'html',crossDomain:true,type:'POST', 
      url:'/php/exec.php',data : {g-product:g-product, g-reason:g-reason}, 
      success:function(html){ 
       console.log(html); 
      }, 
      error:function(j,t,e){ 
       console.log('problem'); 
      } 
     }); 
    }); 
});

你可以試試這個代碼，也許你的文件的js路徑或php路徑存在問題。也許可以糾正一些錯誤。

來源

2013-09-23 09:47:12

如何使按鈕在同一頁面上執行，但不在下一頁

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