<table border="2">
<tr>
<th>Memeber Id</th>
<th>Lastname</th>
<th>Firstname</th>
<th>Birthdate</th>
<th>Gender</th>
<th>Status</th>
<th>Dedication Date</th>
<th>Acceptance Date</th>
<th>Baptism Date</th>
<th>Mother</th>
<th>Father</th>
<th>Decription</th>
</tr>
<?php
$con = mysql_connect("localhost", "root", "");
$er = mysql_select_db("memberdb");
$query = "insert into memberinformation values('$memberid','$lastname','$firstname','$birthdate','$gender','$status','$dedicationdate'
,'$acceptancedate','$baptismdate','$mother','$father','$description')";
$result = mysql_query($query);
$result = mysql_query("SELECT * FROM memberinformation");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
//you need to exit the script, if there is an error
exit();
}
while ($array = mysql_fetch_row($result));
{
echo "<tr>";
echo "<td>" . $row['memberid'] . "</td>";
echo "<td>" . $row['lastname'] . "</td>";
echo "<td>" . $row['firstname'] . "</td>";
echo "<td>" . $row['birthdate'] . "</td>";
echo "<td>" . $row['gender'] . "</td>";
echo "<td>" . $row['status'] . "</td>";
echo "<td>" . $row['dedicationdate'] . "</td>";
echo "<td>" . $row['acceptancedate'] . "</td>";
echo "<td>" . $row['baptismdate'] . "</td>";
echo "<td>" . $row['mother'] . "</td>";
echo "<td>" . $row['father'] . "</td>";
echo "<td>" . $row['description'] . "</td>";
echo "</tr>";
echo "<script> alert('Tama ka.'); </script>";
}
echo "</table>";
mysql_close($con);
?>
我遇到了问题,因为代码没有输出我期望的内容。它应该从成员数据库中的memberinformation表输出数据。如果此代码中有一些缺少的行或者我有一些错误,请帮助我。
答案 0 :(得分:2)
您正在读取值$ row,而数据位于$ array:
while ($array = mysql_fetch_row($result));
...
echo "<td>" . $row['memberid'] . "</td>";
这应该有效:
while ($row = mysql_fetch_row($result)){
...
echo "<td>" . $row['memberid'] . "</td>";
...
}
答案 1 :(得分:2)
你混淆了MySQL和MySQLi库。
将mysqli_connect_errno()
替换为mysql_errno()
以使用MySQL库。 (也适用于mysqli_connect_error()
)
但是不要使用已弃用的MySQL库(mysql_*
函数),这一点非常重要。用MySQLi和PDO替换它。
答案 2 :(得分:1)
您有$array = mysql_fetch_row($result)
,但在表中您正在打印,例如$ row ['memberid']而不是$array['memberid']
。我相信这将是你的问题。
答案 3 :(得分:1)
将$ array = mysql_fetch_row($ result)更改为$ row = mysql_fetch_row($ result)
答案 4 :(得分:0)
mysql_
函数mysql_
和mysqli_
)。请改为选择mysqli_*
。$row = mysql_fetch_row($result)
示例:
<?php
error_reporting(E_ALL);
ini_set("display_errors", 1);
$con = new mysqli('localhost', 'root', '', 'memberdb');
$result = mysqli_query($con, 'SELECT * FROM memberinformation');
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
echo "<script> alert('May Mali Ka.');</script>";
//you need to exit the script, if there is an error
exit();
}
$stmt = $con->prepare('INSERT INTO memberinformation VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)');
$stmt->bind_param('ssssssssssss', $memberid,$lastname,$firstname,$birthdate,$gender,$status,$dedicationdate,$acceptancedate,$baptismdate,$mother,$father,$description);
$stmt->execute();
?>
<style>table, tr, td {border: 1px solid black;}</style>
<table>
<tr>
<th>Memeber Id</th><th>Lastname</th><th>Firstname</th><th>Birthdate</th><th>Gender</th><th>Status</th><th>Dedication Date</th>
<th>Acceptance Date</th><th>Baptism Date</th><th>Mother</th><th>Father</th><th>Decription</th>
</tr>
<?php while($row = $result->fetch_assoc()): ?>
<tr>
<td><?php echo $row['lastname']; ?></td>
<td><?php echo $row['firstname']; ?></td>
<td><?php echo $row['birthdate']; ?></td>
<td><?php echo $row['gender']; ?></td>
<td><?php echo $row['status']; ?></td>
<td><?php echo $row['dedicationdate']; ?></td>
<td><?php echo $row['acceptancedate']; ?></td>
<td><?php echo $row['baptismdate']; ?></td>
<td><?php echo $row['mother']; ?></td>
<td><?php echo $row['father']; ?></td>
<td><?php echo $row['description']; ?></td>
</tr>
<?php endwhile; ?>
</table>