在提交頁面時使用ajax保留選定的複選框

Question

請幫助我使用我的代碼這是我的代碼片段..我使用ajax在提交後獲取我選擇的單選按鈕的數據我希望我的選定複選框保留。

<?php 
$sql = "SELECT * FROM tbl_func"; 
$res = mysql_query($sql) or die(mysql_error()); 
$row = mysql_fetch_array($res); 
$let = $row['letter']; 
?> 
<form action="func.php" method="post"> 
<input onchange="let()" name="letter" type="radio" value="A" <?php if($let == "A") echo "checked";?>/>A 
<input onchange="let()" name="letter" type="radio" value="B" <?php if($let == "B") echo "checked";?>/>B 
<script> 
function let(){ 
    $.ajax({ 
     type:"post", 
     url:"ajax.php", 
     data:{ 
      letter_val:$("[name=letter]:checked").val(), 
      }, 
     success:function(msg){ 
      document.getElementById("div").innerHTML = msg; 
      } 
    }) 
    } 
</script> 
<br /><br /> 
<div id="div"></div> 
<br /> 
<?php if(isset($let)) {?> 
    <input type="submit" name="update" value="update" /> 
<?php }else{?> 
    <input type="submit" name="submit" value="submit" /> 
<?php }?> 
</form> 
<?php 
if(isset($_POST['submit'])){ 
    $letter = $_POST['letter']; 
    $val = $_POST['val']; 
    $val_sel = ""; 
    $val_temp = ""; 
    foreach($val as $val_temp){ 
     $val_sel .= $val_temp.","; 
    } 
    $val_trim = rtrim($val_sel,","); 
    $sql = "INSERT INTO tbl_func (letter,val) VALUES('$letter','$val_trim')"; 
    $res = mysql_query($sql) or die (mysql_error()); 
    if($res) echo "success"; 
    } 
?> 

ajax.php

<?php 
$letter_val = $_POST['letter_val']; 
$sql = "SELECT * FROM tbl_func"; 
$res = mysql_query($sql) or die(mysql_error()); 
$row = mysql_fetch_array($res); 
$val = $row['val']; 
$val_arr = explode(",",$val); 
$val_temp = ""; 

if($letter_val == "A"){ 
?> 
<input type="checkbox" value="1" name="val[]" 
<?php 
foreach($val_arr as $val_temp){ 
    if($val_temp == 1) echo "checked"; 
} 
?>/>1 
<input type="checkbox" value="2" name="val[]" 
<?php 
foreach($val_arr as $val_temp){ 
    if($val_temp == 2) echo "checked"; 
} 
?> 
/>2 
<?php 
} 
if($letter_val == "B"){ 
?> 
<input type="checkbox" value="3" name="val[]" 
<?php 
foreach($val_arr as $val_temp){ 
    if($val_temp == 3) echo "checked"; 
} 
?> 
/>3 
<input type="checkbox" value="4" name="val[]" 
<?php 
foreach($val_arr as $val_temp){ 
    if($val_temp == 4) echo "checked"; 
} 
?> 
/>4 
<?php 
} 
?> 

我使用此代碼，但沒有運氣，請幫助我如何..

func.php。謝謝大家

Answer 1

完成下列步驟，你會明白你的問題的原因：

print_r($val_arr); 

foreach($val_arr as $val_temp){ 
    if($val_temp == 2) echo "checked"; 
    echo $val_temp; 
}