我对PHP非常陌生,我对这个简单任务的错误感到沮丧。我想从SQL数据库导入一个表并将其显示在HTML表中。但是在尝试获取表列名时我一直遇到错误。
虽然我测试了与数据库的连接。
错误是:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Table' at line 1.
上通过其他示例进行了编辑
如果有人可以帮助我,我会很感激。
<?php
error_reporting(-1);
$con = mysqli_connect('localhost', 'user', 'pass');
mysqli_select_db($con, 'database') or die("Could not found" . mysqli_error($con));
$query = ("select * from Table");
$result = mysqli_query($con, $query) or die ( mysqli_error ($con) );
//Print table
echo "<table>";
echo "<tr>";
if($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
//Print headers
foreach($row as $key => $value ){
echo "<td>" . $key . "</td>";
}
echo "</tr>";
}
$result = mysqli_query($con, $query) or die (mysqli_error());
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
echo "<tr>";
while( list($key, $value) = each($row)){
//Print value
echo "<td>" . $value . "</td>";
}
echo "<td>" . $value . "<i class='fa fa-caret-up'></i><i class='fa fa-caret-down'></i></td>";
echo "</tr>";
}
echo "</table>";
?>
答案 0 :(得分:1)
Table
是一个保留关键字,您无法像这样使用它。如果你想从名为users
的表中获取一些数据或者像这样的表,那么查询应该是 -
select * from users