我正在尝试将CSS添加到我的表单中，但不知道如何做到这一点。该表单是在php和MySQL中创建的，在浏览器中看起来像：http://gyazo.com/5d099ead9bd6ea83859a5114b2438748
我需要将文本和下拉列表对齐，使它们在整个过程中相等，并添加一些间距。有人帮过CSS吗？
html当前：
<div class="wrap">
<img src="/images/logo.png" alt="Highdown logo" />
<h1>Entry form</h1>
</div>
css当前：
.wrap {
position: relative;
}
表格是用这个生成的：
if ($event_result = $con->query("SELECT Event.Name FROM event")) {
echo "<form method ='"POST'" action='"save.php'"> ";
while ($row = $event_result->fetch_assoc()) {
echo $row['Name']. ' ';
if ($student_result = $con->query("SELECT Student.Form, Teacher.Form, Student.Forename, Student.Surname, Student_ID " .
"FROM Student, Teacher " .
"WHERE Student.Form = Teacher.Form AND Teacher.Username = '" . $_SESSION['Username'] . "'")) {
if ($student_result->num_rows) {
echo "<select name ='". $row['Name']."'>";
while ($row1 = $student_result->fetch_assoc()) {
echo '<option value="" style="display:none;"></option>';
echo "<option value ='" . $row1['Student_ID'] . "'>" . $row1['Forename'] . ' ' . $row1['Surname'] . "</option>";
}
echo "</select> <br />";
}
}
}
echo '<input type="submit" value ="Submit">';
echo '<input type="reset" value ="Reset">';
echo '<input type="button" value = "Add student" onclick="location.href=''http://localhost/sportsday/addstudent.php''">';
echo '<input type="button" value = "Delete student">';
echo "</form>";
}
使用
<form>
<table>
<tr> //1st Table row
<td></td> //Table column data
<td></td> //table column data
</tr> //1st row ends
<tr> // 2nd Table row
<td></td> //Table column data
<td></td> //table column data
</tr> //2nd row ends
</table>
</form>
这将为您提供更好的表单布局。
我没有尝试，因为我没有数据库
//Query to display all events
if ($event_result = $con->query("SELECT Event.Name FROM event")) {
echo "<form method ='"POST'" action='"save.php'"> ";
echo '<table>';
echo '<tr>';
echo '<td>';
while ($row = $event_result->fetch_assoc()) {
echo $row['Name']. ' ';
echo '</td>';
if ($student_result = $con->query("SELECT Student.Form, Teacher.Form, Student.Forename, Student.Surname, Student_ID " .
"FROM Student, Teacher " .
"WHERE Student.Form = Teacher.Form AND Teacher.Username = '" . $_SESSION['Username'] . "'")) {
if ($student_result->num_rows) {
echo '<td>';
echo "<select name ='". $row['Name']."'>";
while ($row1 = $student_result->fetch_assoc()) {
echo "<option value ='" . $row1['Student_ID'] . "'>" . $row1['Forename'] . ' ' . $row1['Surname'] . "</option>";
}
echo "</select> <br />";
echo '</td>';
echo '</tr>';
}
}
}
echo '</table>';
echo '<input type="submit" value ="Submit">';
echo '<input type="reset" value ="Reset">';
echo '<input type="button" value = "Add student" onclick="location.href=''http://localhost/sportsday/addstudent.php''">';
echo '<input type="button" value = "Delete student">';
echo "</form>";
}
?>
您可以直接在css 中编写
form {
⋮ declare css
}
或命名以形成
form[name="value"]{
⋮ declare css
}
或在表单上添加任何类别或id
#formid{
⋮ declare css
}
.formclass{
⋮ declare css
}
首先，检查数据库。。。可能存在与表格输出无关的另一个问题。
因此，首先删除表标记。。并检查它是否工作？
然后尝试HTML表格标签
否则，给我一个示例数据库.sql文件，并在谷歌驱动器或共享驱动器上完成PHP代码。
这样我就可以检查并确定问题出在哪里了？