將傳遞的值從javascript返回到php

Question

您好我真的是新手，在編程和做一些PHP開發以支持個人承諾。

我寫了一個JavaScript來做一些計算並返回答案。我想從我的php中訪問這個值並打印出來。

以下是代碼組件。

<?php include 'dbconnection.php'; ?> 
<?php 

    $Kg=0.00; 
$Rate=0.00; $MinusBags=0.00; $Amount = 0.00; 
$KgIntoRate=0.00; 
//$Amount = $KgIntoRate - $MinusBags; 



//execute the SQL query and return records 
$result = mysql_query("SELECT EmployeeName FROM employees"); 
$result2 = mysql_query("SELECT SRate FROM scraprates where SYear=YEAR(CURRENT_DATE()) AND SMonth=MONTHNAME(CURRENT_DATE())"); 
$temp1 = mysql_fetch_array($result2); 
$Rate = $temp1['SRate']; 
//$Rate=mysql_fetch_array($result2); 


//fetch the data from the database 
try{ 
while ($row = mysql_fetch_array($result)) { 
    echo "<tr><td>".$row{'EmployeeName'}."</td>"; 
    echo "<td><input type='text' name='Kg' id='Kg' onChange='CalcKgIntoRate();'/></td>"; 
    echo "<td>".$Rate."</td>"; 


// $KgIntoRate = $_GET['php_KgIntoRate']; 
    //echo "<td>".$_GET['php_KgIntoRate']."</td>"; 
    echo "<td>".$KgIntoRate."</td>"; 

    echo "<td><input type='text' name='MinusBags'/></td>"; 
    echo "<td>".$Amount."</td></tr>"; 
} 
} 
catch(Exception $e) 
{echo "error".$e;} 
?> 

<script type="text/javascript"> 
//$Kg= $('#Kg').val(); 
function CalcKgIntoRate(){ 
var Kg=document.getElementById('Kg').value; 
var php_rate = "<?php echo $Rate; ?>"; 
var php_KgIntoRate = "<?php echo $KgIntoRate; ?>"; 
//document.write(Kg); 
php_KgIntoRate=Kg*php_rate; 
return php_KgIntoRate; 
} 
</script> 

<?php 
//function CalcKgIntoRate(){ 
//$KgIntoRate = $Kg * $Rate; 
//echo "<td>".$KgIntoRate."</td>"; 
//} 


//close the connection 
mysql_close($dbhandle); 
?> 

我想要做的是這樣的。名稱和比率來自兩個數據庫表。我想根據輸入的Kg（在文本字段的變化事件）計算KgIntoRate，並在Kg * Rate字段中顯示值。

我看到我需要使用Ajax。但不知道該怎麼做。感謝一些有關代碼的幫助。如果它可以幫助你

檢查：

Answer 1

你可以做到這一點wihtout阿賈克斯，如果你的計算是在JavaScript處理。

例如 的變化是PHP代碼：

$i = 0; // define outside the loop 

echo "<td><input type='text' name='Kg' id='Kg".$i."' onChange='CalcKgIntoRate(".$i.");'/></td>"; 
echo "<td id='rate".$i."'>".$KgIntoRate."</td>"; 
echo "<td id='kg_into_rate".$i."'>".$Amount."</td></tr>"; 
$i++; 

JS代碼：

function CalcKgIntoRate(i) { 
    var kg = document.getElementById('Kg'+i).value; 
    var rate = document.getElementById('rate'+i).innerHTML; 
    var kg_into_rate = document.getElementById('kg_into_rate'+i); 

    // here is your calculation login 
    // suppose you store the calculation in an another variable 
    var calVal = parseFloat(kg*eval(rate)); 
    kg_into_rate.innerHTML = calVal; 
} 

Answer 2

我想在你的榜樣你的思念是剛好路過值返回到輸入字段。

function CalcKgIntoRate(){ 
var Kg=document.getElementById('Kg').value; 
var php_rate = "<?php echo $Rate; ?>"; 
var php_KgIntoRate = "<?php echo $KgIntoRate; ?>"; 
//document.write(Kg); 
php_KgIntoRate=Kg*php_rate; 
//return php_KgIntoRate; 

// here! instead of returning the value, set it as the value of your Input 
document.getElementById('Kg').value = php_KgIntoRate;} 

，或者你可以簡單地做在PHP計算....

$result1 = $Rate * $KgIntoRate; 
echo "<tr><td>".$row{'EmployeeName'}."</td>"; 
echo "<td><input type='text' name='Kg' id='Kg' value='".$result1."'/></td>"; 
echo "<td>".$Rate."</td>";