php - php表单复选框='$变量'

Question

我似乎无法完成这件事。我从表中获取数据，如果列中的值 = y，则必须选中复选框，如果列中的值 = n，则不得选中复选框，所有复选框都保持选中状态，即使只有第一行的列 = y

<?php
session_start();
    include 'datalogin.php';
    $sel_course_id = 2;
    $sel_date_1    = '2012-05-21';
    $sel_date_2    = '2012-05-22';
    echo "<form action='' method='post'>";
    echo "<strong>Completed the course</strong>";

    $sql1 = "SELECT * FROM trn WHERE course_id = '$sel_course_id' AND date_1 = '$sel_date_1' and date_2 = '$sel_date_2'";
    $res1 = mysql_query($sql1);
    while ($row1 = mysql_fetch_assoc($res1)) {
        $rowid1           = $row1['id'];
        $uid1              = $row1['usr_id'];
        $certificate_name1 = $row1['certificate_name'];
        $course_completed1 = $row1['course_completed'];

            $sql2 = "SELECT * FROM ex_usrs WHERE id = '$uid1'";
            $res2 = mysql_query($sql2);
            while ($row2 = mysql_fetch_assoc($res2)) {
                $fn2 = $row2['firstname'];
                $ln2 = $row2['lastname'];
            } 
            $checked2 = "";
            if ($course_completed1 == y) {
                $checked2 = "checked";
            } 
            if ($course_completed1 == n) {
                $checked2 = "";
            } 
            echo "<input name='question[$rowid1][]' type='checkbox'  checked='$checked2' value='1' />$fn2 $ln2";
echo "</br>";
   } 
    echo "</form>";
?>

Answer 1

对字符串使用引号：

if ($course_completed1 == "y") {
    $checked2 = "checked";
} 
if ($course_completed1 == "n") {
    $checked2 = "";
} 

Answer 2

checked='$checked2'应该是 plain $checked2，它是一个布尔属性。

（除非您正在编写 XHTML，在这种情况下，它也应该'checked="checked"在您 test 之后被赋予该值if ($course_completed1 == "y") {）。

您的测试还应该检查字符串 ( "y") 而不是常量 ( y)。

Answer 3

如前所述，您需要引用y并n比较字符串值。但即使只是拥有checked=""也会导致您的复选框被选中。试着做这样的陈述：

if ($course_completed1 == "y") {
    $checked2 = "checked='checked'";
} 
elseif ($course_completed1 == "n") {
    $checked2 = "";
}

echo "<input name='question[$rowid1][]' type='checkbox' $checked2 value='1' />$fn2 $ln2";