如何將值傳遞到PHP以獲取表作爲輸出？

-1

<script type="text/javascript"> 
    function OnSliderChanged (slider) { 
     var sliderValue = document.getElementById (slider.id + "Value"); 
     sliderValue.innerHTML = slider.value; 
    } 

    function Init() { 
     var slider = document.getElementById ("slider1"); 
     OnSliderChanged (slider); 
     var slider = document.getElementById ("slider2"); 
     OnSliderChanged (slider); 
    } 
</script> 
0<input type="range" id="slider1" min="0" max="100" step="1" style=width:50% onchange="OnSliderChanged (this)" />100 
<span id="slider1Value" class="sliderValue"></span> 
<br /><br /><br /> 
0<input type="range" id="slider2" min="0" max="100" step="1" style=width:50% onchange="OnSliderChanged (this)" />100 
<span id="slider2Value" class="sliderValue"></span> 
<br> 
<input type="submit" onmouseover="this.style.cursor=\'pointer\';" OnMouseOut="this.style.cursor=\'default\';"> 


<?php 
if (isset($_GET['value=20'&&'value1=40'])) 
{ 
$username = 'root'; 
$password = ''; 
$hostname = 'localhost'; 
$dataname = mysql_connect($hostname,$username,$password) or die("Connection failed: "); 
$selected = mysql_select_db('testdb',$dataname) or die("Could not select table1"); 
$result = mysql_query("SELECT * FROM `table1`"); 
$last = ''; 
echo "<table> 
    <tr> 
     <th>id</th> 
     <th>name</th> 
    </tr>"; 
while($row = mysql_fetch_array($result)) 
{ 
echo"<tr> 
     <td>".$row['0']."</th> 
     <td>".$row['1']."</td> 
    </tr>"; 
    } 
mysql_close($dataname); 
} 
else 
{ 
echo " No Values Found"; 
} 
?>

我想獲取輸入值到PHP應該顯示錶。但我沒有得到表作爲結果得到「沒有發現值」在初始頁面本身。這是新的PHP n JavaScript ..例如，我需要得到的值，當20 n 40輸入值作爲輸入...可以有人請幫助我..？如何將值傳遞到PHP以獲取表作爲輸出？

來源

2014-12-30 Siva Shanmugam

'isset（$ _GET [ '值= 20' && '值1 = 40']）'？ –

將其修正爲「if（（isset（$ _ POST ['value']）&&（$ _POST ['value'] =='20'））&&（isset（$ _ POST ['value1']）&&（$ _POST ['value1'] =='40'）））「還沒有越來越.. –

給輸入一個名字。 –

你必須使用HTML形式name屬性才能訪問他們在服務器端..

0<input type="range" name="value" id="slider1" min="0" max="100" step="1" style=width:50% onchange="OnSliderChanged (this)" />100 
<span id="slider1Value" class="sliderValue"></span> 
<br /><br /><br /> 
0<input type="range" name="value1" id="slider2" min="0" max="100" step="1" style=width:50% onchange="OnSliderChanged (this)" />100 
<span id="slider2Value" class="sliderValue"></span> 
<br> 
<input type="submit" onmouseover="this.style.cursor=\'pointer\';" OnMouseOut="this.style.cursor=\'default\';">

而且使窗體方法到後與method="post"

來源

2014-12-30 06:35:11

'value'應該是'20'，'value1'應該是'40'。所以在你的表單中正確設置'value'屬性.. –

'」method =「POST」>' –

如何將值傳遞到PHP以獲取表作爲輸出？

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