使用Ajax發佈該值

Question

我正在嘗試發佈輸入名稱=「sub2」的值，以在我的輸入=「sub3」中處理我的查詢。我使用Ajax，但輸入名稱=「sub3」仍爲空。 sub2的值來自下拉選項文本，如果選項文本值爲Supplies它只獲得3個第一個字母的電源，所以現在sub2的值是「SUP」。這個值我需要張貼在我的sub3文本框中運行我的查詢。請幫助？

這是小提琴例如http://jsfiddle.net/xqGLS/3/

Code.php

<?php 
    $resultcode = $mysqli->query("SELECT category, id, maincode FROM category GROUP BY id ORDER BY maincode"); 
    $code = ''; 
    while($row = $resultcode->fetch_assoc()) 
     { 
     $code .= '<option value = "'.$row['maincode'].'">'.$row['category'].'</option>'; 
     } 
?> 
<form action="<?php $_SERVER['PHP_SELF']?>" method="POST"> 
<br/> 

<label>Category</label> 
<select name="maincode" style="text-transform:uppercase;" onchange = "GetChangedValue(this);"> 
<option value=""></option> 
<?php echo $code; ?> 
</select> 

</br> 

<label>Sub1</label> 
<input type="" name="sub1" id="sub1" value="" readonly style="width:45px;text-transform:uppercase;"> 

<script> 
$('[name="maincode"]').change(function() { 
    $('[name="sub1"]').val($(this).val()); 
    var input = $('[name="sub2"],[name="sub3"]'), 
    input1 = $('[name="sub2"]'), 
    input2 = $('[name="sub3"]'), 
    input3 = $('[name="equal"]'); 
    input.change(function() { 
    input3.val(input1.val() + input2.val()); 
}); 
}); 
</script> 

<script> 
function GetChangedValue(e) { 
var value = e.options[e.selectedIndex].text; 
var elem = document.getElementById("sub2"); elem.value = value.substring(0,3); 
} 
</script> 

<label>Sub2</label> 
<input name="sub2" id="sub2" value="" style="width:35px;text-transform:uppercase;" readonly> 

<label>Sub3</label> 
<input id="sub3" name="sub3" style="width:35px;text-transform:uppercase;" value=''> 

<script type="text/javascript"> 
$('#sub2').change(function(){ 
$.ajax({ 
url : 'check.php', 
data :{mainlist_id : $(this).val()}, 
dataType:'html', 
type:'POST', 
success:function(data){ 
$('#sub3').html(data); 
} 
}); 
}); 
</script> 

<input id="equal" name="equal" value="" style="width:60px;text-transform:uppercase;" type="hidden"> 
<input type="submit" name=""> 
</form> 

Check.php

<?php 
    $mysqli = new mysqli("localhost", "root", "", "2015"); 
    $sub = $mysqli->real_escape_string($_POST["mainlist_id"]); 

$result2 = $mysqli->query("SELECT * FROM code 
WHERE sub1code LIKE '$sub-___' ORDER BY sub1code"); 

while($row = $result2->fetch_assoc()) 
    { 
    $value = $row['sub1code']; 
    } 
    $first = substr($value, 0, 4); 
    echo $first; 
    $last = substr($value, -3); 
    $i="0"; 
    while($i<=$last) 
    { 
    $i++; 
    } 
    $value2=strlen($i); 
    echo $first; 
    if($value2==1) 
    { 
    echo "00".$i; 
    } 
    elseif($value2==2) 
    { 
    echo "0".$i; 
    } 
    else 
    { 
    echo $i; 
    } 
    ?> 

Answer 1

我想你想要這樣的東西。只是對你的代碼做一點改動，但它應該可以工作。

Code.php

<script type="text/javascript"> 
function GetChangedValue(str) { 
    if (str==""){ 
     document.getElementById("sub3").innerHTML=""; 
     return; 
    } 
    if (window.XMLHttpRequest) { 
     // code for IE7+, Firefox, Chrome, Opera, Safari 
     xmlhttp=new XMLHttpRequest(); 
    } else { 
     // code for IE6, IE5 
     xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); 
    } 
    xmlhttp.onreadystatechange=function(){ 
     if (xmlhttp.readyState==4 && xmlhttp.status==200){ 
      document.getElementById("sub3").value = xmlhttp.responseText; 
     } 
    } 
    xmlhttp.open("GET","check.php?q="+str,true); 
    xmlhttp.send(); 
} 
</script> 

<select name="category" id="category" onchange="GetChangedValue(this.value);"> 
<option value=""></option> 
<?php 
$mysqli = new mysqli("localhost", "root", "", "2015"); 
$result2 = $mysqli->query("SELECT * FROM category ORDER BY maincode"); 
while($row = $result2->fetch_assoc()) 
{ 
$ok = $row['maincode']; 
echo "<option value=".$row['maincode'].">".$row['category']."</option>"; 
}        
?> 
</select> 


<input type="" name="sub3" id="sub3" value="" readonly style="width:85px;text-transform:uppercase;"> 

Check.php

<?php 
if (isset($_GET["q"])) { 
$q=$_GET["q"]; 
$mysqli = new mysqli("localhost", "root", "", "2015"); 

$sub = substr($q, 0, 4); 

$result2 = $mysqli->query("SELECT * FROM code 
WHERE sub1code LIKE '$sub-___' ORDER BY sub1code"); 

while($row = $result2->fetch_assoc()) 
    { 
    $value = $row['sub1code']; 
    } 
    //$first = substr($value, 0, 3); 
    echo $sub; 
    $last = substr($value, -3); 
    $i="0"; 
    while($i<=$last) 
    { 
    $i++; 
    } 
$value2=strlen($i); 
if($value2==1) 
{ 
echo "-"."00".$i; 
} 
elseif($value2==2) 
{ 
echo "-"."0".$i; 
} 
else 
{ 
echo "-".$i; 
} 
} 
?> 

Answer 2

有2個問題：

1. 應該首先添加事件而不是其他事件。如果你喜歡 這個，當maincode改變時，sub1和sub2已經改變，但是現在 sub1和sub2沒有預期的改變事件。
2. 輸入的值被腳本更改無法觸發更改 事件。