忽略禁用的复选框

Question

我这里有一个代码，当您单击 All link 时，它将选中所有复选框，当 None 它将取消选中所有复选框。

但是我有一个PHP条件，当$pta_fee == $pta_fee_trans复选框将被禁用，但是当我点击所有 link，禁用复选框似乎已被选中。

如果我选中全部 link，如何忽略禁用复选框？

<div class="sub_profile right">
<p class="sub_content_text" style='margin-left: 25px;'>
    <a href="javascript:selectToggle(true, 'drawing');" id="show">All</a> | <a href="javascript:selectToggle(false, 'drawing');" id="hide">None</a>
     MISCELLANEOUS FEES:
</p>
<?php
    if($pta_fee == $pta_fee_trans)
    {
?>
<p class="sub_content_text" style='margin-left: 30px;'>
    <input type='checkbox' value="<?php echo $pta_fee; ?>" disabled>
    PTA Fee : &#8369; <?php echo $pta_fee; ?></p>
<?php
    }
    else
    {
?>
<p class="sub_content_text" style='margin-left: 30px;'>
    <input type='checkbox' name='draw[]' value="<?php echo $pta_fee; ?>" id="required-checkbox1" onClick="CheckIfChecked(this.id)">
    PTA Fee : &#8369; <?php echo $pta_fee; ?></p>
<?php
    }
    if($maintenance_fee == $maintenance_fee_trans)
    {
?>
<p class="sub_content_text" style='margin-left: 30px;'>
    <input type='checkbox' value="<?php echo $maintenance_fee; ?>" disabled>
    Maintenance Fee : &#8369; <?php echo $maintenance_fee; ?></p>
<?php
    }
    else
    {
?>
<p class="sub_content_text" style='margin-left: 30px;'>
    <input type='checkbox' name='draw[]' value="<?php echo $maintenance_fee; ?>" id="required-checkbox2" onClick="CheckIfChecked(this.id)">
    Maintenance Fee : &#8369; <?php echo $maintenance_fee; ?></p>
<?php
    }
    if($id_school == $id_school_trans)
    {
?>
<p class="sub_content_text" style='margin-left: 30px;'>
    <input type='checkbox' value="<?php echo $id_school; ?>" disabled>
    School ID : &#8369; <?php echo $id_school; ?></p>
<?php
    }
    else
    {
?>
<p class="sub_content_text" style='margin-left: 30px;'>
    <input type='checkbox' name='draw[]' value="<?php echo $id_school; ?>" id="required-checkbox3" onClick="CheckIfChecked(this.id)">
    School ID : &#8369; <?php echo $id_school; ?></p>
<?php
    }
    if($electricity == $electricity_trans)
    {
?>
<p class="sub_content_text" style='margin-left: 30px;'>
    <input type='checkbox' value="<?php echo $electricity; ?>" disabled>
    Electricity : &#8369; <?php echo $electricity; ?></p>
<?php
    }
    else
    {
?>
<p class="sub_content_text" style='margin-left: 30px;'>
    <input type='checkbox' name='draw[]' value="<?php echo $electricity; ?>" id="required-checkbox4" onClick="CheckIfChecked(this.id)">
    Electricity : &#8369; <?php echo $electricity; ?></p>
<?php
    }
?>
<div id="sub_profile_cont">
    <div class="sub_profile left">
        <p class="block_cont left">
            <div id = "submit-button-container" style="display:none;">
                <input class="action_btn" type="submit" name="submit" id="pay_btn" value="COMPUTE" onClick="setUpdateAction();"/>
            </div>
            <b style="display: none";><input class="action_btn" type="submit" name="submit" id="pay_btn" value="COMPUTE" onClick="setUpdateAction();"/></b>
        </p>
    </div>
</div>
</div>

--这是检查所有复选框的javascript代码：

<script>
function selectToggle(toggle, form) {
    var myForm = document.forms[form];
    for( var i=0; i < myForm.length; i++ ) { 
    if(toggle) {
         myForm.elements[i].checked = "checked";
    } 
    else {
         myForm.elements[i].checked = "";
         }
    }
}

Answer 1

试试这个：

<script>
function selectToggle(toggle, form) {
    var myForm = document.forms[form];
    for( var i=0; i < myForm.length; i++ ){ 
        if(myForm.elements[i].disabled != true){
            if(toggle){
                 myForm.elements[i].checked = "checked";
            } 
            else{
                 myForm.elements[i].checked = "";
            }
        }   
    }
}

忽略禁用的复选框

Ignore checkbox that is disable

javascript

php

checkbox

disabled-input