Question

页面中有两个表。一个是星期五，另一个是星期六。我想从表中设置单选按钮选择的选择限制。用户不能从星期五表中的4个插槽中选择3个以上的插槽。如何为表调用javascript函数？我怎样才能做到这一点？这是我的PHP代码和javascript函数。

这个javascript函数适用于两个表。当我从任何表格中选择3个单选按钮时，它会显示“你不能添加超过3个”。但我希望这只适用于星期五的表而不是星期六。

PHP代码：

              for($time=0; $time<4; $time++){
                $q.= "<td colspan='1' valign='top' width='25%'>";
                    $q.= "<table id='ash'>";
                    $q.= "<tr id='grey'>";
                    if($day==1){
                        if($time==0){
                            $q.= "<th width='50%'></th></tr>";
                        }else if($time==1){
                            $q.= "<th width='50%'></th></tr>";
                        }else if($time==2){
                            $q.= "<th>Reg.</th>";
                            $q.= "<th width='50%'>2:30 pm - 5:00 pm</th>";
                            $q.= "<th>Sec</th><th>Remaining</th></tr>";
                        }else{
                            $q.= "<th>Reg.</th>";
                            $q.= "<th width='50%'>5:00 pm - 7:30 pm</th>";
                            $q.= "<th>Sec</th><th>Remaining</th></tr>";
                        }
                    } 
                  else{
                        $q.= "<th>Reg.</th>";
                        if($time==0){
                            $q.= "<th width='50%'>8:30 am - 11:00 am</th>";
                        }else if($time==1){
                            $q.= "<th width='50%'>11:00 am - 1:30 pm</th>";
                        }else if($time==2){
                            $q.= "<th width='50%'>2:30 pm - 5:00 pm</th>";
                        }else{
                            $q.= "<th width='50%'>5:00 pm - 7:30 pm</th>";
                        }
                        $q.= "<th>Sec</th><th>Remaining</th></tr>";
                    }
                    $query1 = "SELECT tab1.*, tab2.course_name, tab2.course_type FROM routine AS tab1, course_info AS tab2"; 
                    $result = mysql_query($query1);
                    if ($result && mysql_num_rows($result) > 0) {
                        while($row = mysql_fetch_assoc($result)){
                            $course = $row["course_code"];
                            $sec = $row["section"];
                            $type = $row["course_type"];
                            if(course_completed($std_id, $course)){
                                if(already_selected($course, $sec, $std_id, $year, $sem)){
                                    $q.= "<tr bgcolor='#00FF33'><td><input type='radio' id='course_".$day."_".$time."' name='".$day.$time."' value='".$course.",".$sec.",".$type."' checked='checked' onclick='changeRadioValue(\"".$day."_".$time."\")' /></td>";
                                }
                                else{
                                    $q.= "<tr bgcolor='#00FF33'><td><input type='radio' id='course_".$day."_".$time."' name='".$day.$time."' value='".$course.",".$sec.",".$type."' onclick='changeRadioValue(\"".$day."_".$time."\")' /></td>";
                                }
                            }

                       else{

                                if(already_selected($course, $sec, $std_id, $year, $sem)){
                                    $q.= "<tr><td><input type='radio' id='course_".$day."_".$time."' name='".$day.$time."' value='".$course.",".$sec.",".$type."' checked='checked' onclick='changeRadioValue(\"".$day."_".$time."\")' /></td>";
                                }else{
                                    $q.= "<tr><td><input type='radio' id='course_".$day."_".$time."' name='".$day.$time."' value='".$course.",".$sec.",".$type."' onclick='changeRadioValue(\"".$day."_".$time."\")' /></td>";
                                }
                            }

JavaScript函数

$("input[type='radio']").change(function(){
    var count = $("input[type='radio']:checked").length;
    if(count>3){
        $(this).prop('checked', false);
      alert("You cannot add more than 3");
    }
});

Answer 1

试试这个：

<?php
if(condition) {
    ?>
    <script>//JS Code</script>
    <?php
}
?>

当您加载页面时，如果条件为true，脚本标记之间的js代码将运行，您还可以链接另一个js文件而不是编写代码。

Answer 2

解决了我的问题。我只是改变了我的JS功能

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<script type="text/javascript">

    //table fri
    $("#fri input[type='radio']").change(function(){
        var count = $("#fri input[type='radio']:checked").length;
        if(count>3){
            $(this).prop('checked', false);
          alert("You cannot add more than 3");
        }
    });

    //table sat
    $("#sat input[type='radio']").change(function(){
        var count = $("#sat input[type='radio']:checked").length;
        if(count>3){
            $(this).prop('checked', false);
          alert("You cannot add more than 3");
        }
    });

    //refresh selection
    $(".refresh").click(function(){
        $("input[type='radio']").prop('checked', false);
    });

如何在php中使用javascript函数？

2 个答案: