php - 具体mysql行到php代码

Question

使用下面的代码，我必须向数据库中的行表报告。

但是除了我需要的行之外，还显示包含 php 代码中的部分(noName)。

我尝试只保留 echo 标签

echo $row[0];

但我收到错误

Notice: Undefined offset: 1 in test.php on line 16

MYSQL

ID            |          Team
------------------------------
1                        test1
2                        test2
3                        test3

PHP

<?php
$connection = mysqli_connect("YourHost","user","password","dbName") or die('connection to DB failed');
$query = mysqli_query($connection,"SELECT team FROM s1");
while ($row = mysqli_fetch_array($query,MYSQLI_NUM)) {
?>

<table>
<tr>
    <td><p> team name <?php if(isset($row[0])){
                echo $row[0];
            }else{
                echo 'noName';
            } ?></p></td>
</tr>
<tr>
    <td><p> team name <?php if(isset($row[1])){
                echo $row[1];
            }else{
                echo 'noName';
            } ?></p></td>
</tr>
<?php } ?>

php页面输出:

test1

noName

test2

noName

Answer 1

您仅从数据库中选择 1 个“团队”字段。 因此，在 PHP 代码中，如果您编写 $row[0]，则与您编写 $row['team'] 是相同的。

$row[1] 所以是空的。

如果你想打印所有团队的名称，你必须编写以下代码:

...
echo "<table>";

while ($row = mysqli_fetch_array($query,MYSQLI_NUM)) {
?>


<tr>
    <td><p> team name <?php echo $row[0]; ?></p></td>
</tr>

我会这样写你的代码:

<?php

$connection = mysqli_connect("YourHost","user","password","dbName") or die('connection to DB failed');
$query = mysqli_query($connection,"SELECT team FROM s1");

if($query) {

    echo "<table>";

    while ($row = mysqli_fetch_array($query,MYSQLI_NUM)) {
        echo "<tr>";
        echo "<td><p> team name " . $row['team'] . "</p></td>";
        echo "</tr>";
    }

    echo "</table>";
}

?>