我将数据库导入到PHPmyadmin中,在此数据库中运行良好,但是当我尝试显示时
我的php文件,我不正确,唯一看到的是该字段的顶部。
hier是我的应用程序
<?php
DEFINE('db_name', 'root');
DEFINE('db_pass', 'root');
DEFINE('db_host', 'localhost');
DEFINE('db_database','SCV_DB');
$dbcon =mysqli_connect("localhost", "root", "root", "SCV_DB");
if (mysqli_connect_error()) {
die('Connect Error ('.mysqli_connect_errno().') '.mysqli_connect_error());
}
$color=' well come to mysql';
echo 'Connected successfully<br>';
echo 'M. Mori ' . $color.'<br>';
mysqli_set_charset($dbcon,"utf8");
echo "<table><tr><th>SECTION</th><th>NAME</th><th>DATE</th><th>COUNTRY</th><th>PREIS</th></tr>";
$consult="SELECT * FROM TABLE 1";
$result=mysqli_query($dbcon,$consult);
while($row = mysqli_fetch_row($result)) {
echo "<tr><td>";
echo $row[0]."</td><td>";
echo $row[1]."</td><td>";
echo $row[2]."</td><td>";
echo $row[3]."</td><td>";
echo $row[4]."</td></tr>";
}
echo "</table>";
?>
答案 0 :(得分:0)
我讨厌mysqli的程序样式,应该使用here you can find more更有用的面向对象样式。
无论如何!我在玩程序性连接,看到您的问题希望它不是重复的:)
所以这是我正在玩的连接文件!您可以使用自己的连接。
$host = '127.0.0.1';
$db = 'SCV_DB';
$user = 'root';
$pass = '';
$port = '3308'; //Or 3307
$charset = 'utf8mb4';
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$dbcon = mysqli_connect($host, $user, $pass, $db, $port, $charset);
if ($dbcon ->connect_error) {
die("Connection failed: " . $dbcon ->connect_error);
}
这是带有prepare语句的查询(在更新和插入数据时,prepare语句可避免您进行sql注入操作)
$sql = 'SELECT * FROM YourTableName';
$stmt = mysqli_prepare($dbcon ,$sql);
if(mysqli_stmt_execute($stmt)){
$getResult = mysqli_stmt_get_result($stmt);
while($rows = mysqli_fetch_assoc($getResult)){
echo "id: ".$rows['id'];
echo "<br>";
echo "First name: ".$rows['first_name'];
echo "<br>";
echo "Last name: ".$rows['last_name'];
echo "<br>";
}
}
mysqli_stmt_close($stmt);
答案 1 :(得分:0)
<?php
DEFINE('db_name', 'root');
DEFINE('db_pass', 'root');
DEFINE('db_host', 'localhost');
DEFINE('db_database','SCV_DB');
$dbcon =mysqli_connect("localhost", "root", "root", "SCV_DB");
// Check connection
if (!$dbcon) {
die("Connection failed: " . mysqli_connect_error());
}
echo "Connected successfully";
$color=' well come to mysql';
echo 'Connected successfully<br>';
echo 'M. Mori ' . $color.'<br>';
mysqli_set_charset($dbcon,"utf8");
?>
<table>
<tr>
<th>SECTION</th>
<th>NAME</th>
<th>DATE</th>
<th>COUNTRY</th>
<th>PREIS</th>
</tr>
<?php
$consult="SELECT * FROM TABLE 1";//Please Don't these name for table name
$result=mysqli_query($dbcon,$consult);
while($row=mysqli_fetch_row($result)):?>
<tr>
<td><?php echo $row[0];?></td>";
<td><?php echo $row[1];?></td>";
<td><?php echo $row[2];?></td>";
<td><?php echo $row[3];?></td>";
<td><?php echo $row[4];?></td>";
</tr>
<?php endwhile;?>
</table>
?>
答案 2 :(得分:0)
答案 3 :(得分:0)
我应该在第一篇文章中更好地解释我的问题。好吧,这个文件是一个excel file.xls,我使用Mac,因此要在phpmyadmin中使用它,首先我应该将其转换为CSV文件而不是导入它,这没什么问题,因为我可以在phpmyadmin中完美地看到数据库,但是当我尝试通过php在浏览器中显示它,
得到真实。
答案 4 :(得分:0)
仅重命名表即可解决问题。