Question

我遇到了一些代码问题，所以我从头开始重新开始，但是现在我的原始代码工作的东西在这里工作并不能解决原因。当我从下拉框中选择onchange函数时，它应该触发我的重载函数，没有任何反应。这是我的代码：

<!DOCTYPE html>
<?php
require 'config.php';  // Database connection
//////// End of connecting to database ////////
?>
<html>
<head>
<SCRIPT language=JavaScript>
    function reload(form)
    {
        var val=form.year1.options[form.year.options.selectedIndex].value;
        self.location='spt.php?year1='+val;
    }
</script>
</head>
<body>
<div>
    <?Php
    @$year1=$_GET['year1'];
    @$team1=$_GET['team1'];

    $quer1="SELECT DISTINCT year FROM PlayerRegSeason ORDER BY year";
    $quer2="SELECT DISTINCT team FROM PlayerRegSeason WHERE year=$year1 ORDER BY team";
    $quer3="SELECT fname, lname FROM PlayerRegSeason WHERE year=$year1 and team ='$team1'";

    echo "<form method=post name=f1 action ='searchpageresultsdd.php'>";

    echo "<select name ='year1' onchange=\"reload(this.form)\"><option value=''>Select Year</option>";
    foreach ($dbo->query($quer1) as $row1){
        if($row1['year']==@$year1){echo "<option selected value='$row1[year]'>$row1[year]</option>"."<BR>";}
        else{echo "<option value='row1[year]'>$row1[year]</option>";}
    }
    echo "</select>";
    ?>
</div>
</body>
</html>

Answer 1

由于name代码的<select>属性为year1，因此语法

var val=form.year1.options[form.year.options.selectedIndex].value;

应更改为此

var val=form.year1.options[form.year1.options.selectedIndex].value;

Php select onchange似乎并没有起作用

1 个答案: