我正在尝试将CSS添加到我的表单中,但不确定如何执行此操作。表单在php和MySQL中创建,在浏览器中看起来像:http://gyazo.com/5d099ead9bd6ea83859a5114b2438748
我需要将文本和下拉对齐,以便它们在整个过程中保持平等并添加一些间距。有人帮忙用CSS吗?
html目前:
<div class="wrap">
<img src="/images/logo.png" alt="Highdown logo" />
<h1>Entry form</h1>
</div>
目前css:
.wrap {
position: relative;
}
表格是用以下方式制作的:
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'] . ' ' . $row1['Surname'] . "</option>";
}
echo "</select> <br />";
}
}
}
echo '<input type="submit" value ="Submit">';
echo '<input type="reset" value ="Reset">';
echo '<input type="button" value = "Add student" onclick="location.href=\'http://localhost/sportsday/addstudent.php\'">';
echo '<input type="button" value = "Delete student">';
echo "</form>";
}
答案 0 :(得分:2)
使用
<form>
<table>
<tr> //1st Table row
<td></td> //Table column data
<td></td> //table column data
</tr> //1st row ends
<tr> // 2nd Table row
<td></td> //Table column data
<td></td> //table column data
</tr> //2nd row ends
</table>
</form>
这将为您提供更好的表单布局。
答案 1 :(得分:1)
这应该工作我没有尝试,因为我没有数据库
//Query to display all events
if ($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>";
}
?>
答案 2 :(得分:0)
你可以直接用css写
form {
⋮ declare css
}
或给表格命名
form[name="value"]{
⋮ declare css
}
或在表单上添加任何类或ID
#formid{
⋮ declare css
}
.formclass{
⋮ declare css
}
答案 3 :(得分:0)
首先,检查一下你的数据库...... 可能存在与表格输出无关的另一个问题。
那么,首先删除Table Tag ..并检查它是否正常工作?
然后尝试HTML TABLE TAG
否则给我示例数据库.sql文件并在谷歌驱动器或共享驱动器上完成PHP代码。
这样我可以检查并确定问题在哪里?