我在MySQL中有一个数据库,但该表没有提供正确的信息。例如,当我运行我的代码时,我在excel中的表以及现在在MySQL中的表自动出现在我的网页上,但数据都是错误的,几乎都在错误的位置。唯一正确的部分是玩家编号和名称位于正确的位置,但即便如此,整个名称甚至都没有显示。然后位置列是所有随机字母,然后高度是重量,重量是年龄,依此类推。
另外,我的excel表当然有正确的所有东西。我只是不确定为什么在创建这个数据库时会输入错误的信息
这是我的整个PHP代码:
<?php
$db_conn = mysqli_connect("localhost", "root", "");
if (!$db_conn)
die("Unable to connect: " . mysqli_connect_error()); // die is similar to exit
mysqli_query($db_conn, "CREATE DATABASE IF NO EXISTS bearsroster;");
if (mysqli_query($db_conn, "CREATE DATABASE bearsroster;"))
echo "Database ready<br>";
else
echo "Unable to create database: " . mysqli_error($db_conn) . "<br>";
mysqli_select_db($db_conn, "bearsroster");
$cmd = "CREATE TABLE sportDatabase ( number int(2) NOT NULL PRIMARY KEY,
playerName char(50),
position varchar(2),
height char(3),
weight int(3),
age int(2),
experience int(2),
collegeName char(60)
);"; //Creates the table
mysqli_query($db_conn, $cmd);
$cmd = "LOAD DATA LOCAL INFILE 'bearsRoster.csv' INTO TABLE sportDatabase FIELDS TERMINATED BY ',';";
mysqli_query($db_conn, $cmd);
echo "<h1>football database contents</h1>";
$cmd = "SELECT * FROM sportDatabase";
$records = mysqli_query($db_conn, $cmd);
echo( "<table border = 'black' align:'center'>
<tr>
<th>Number</th>
<th>Player Name</th>
<th>Position</th>
<th>Height</th>
<th>Weight</th>
<th>Age</th>
<th>Experience</th>
<th>College Name</th>
</tr>" . PHP_EOL );
//Prints out the table
while($row = mysqli_fetch_array($records)){
echo( "<tr>
<td id = 'yellow'>" . $row['number'] . "</td>
<td id = 'red'>" . $row['playerName'] . "</td>
<td id = 'red'>" . $row['position'] . "</td>
<td id = 'blue'>" . $row['height'] . "</td>
<td id = 'blue'>". $row['weight'] . "</td>
<td id = 'blue'>". $row['age'] . "</td>
<td id = 'pink'>". $row['experience'] . "</td>
<td id = 'pink'>". $row['collegeName'] . "</td>
</tr>". PHP_EOL );
}