Question

我确信我的表中有两行数据符合where子句的条件，但我的编码只输出（回显）一行。我究竟做错了什么？

<?php


$connect= mysqli_connect("localhost", "root", "", "friends_list")
or die('error connecting with the database');

$query= "SELECT * FROM people WHERE age=19";

$result= mysqli_query($connect, $query)
or die('error querying the database');

$row= mysqli_fetch_array($result);

while ($row = mysqli_fetch_array($result))
{
echo $row['first_name'] . " " . $row['last_name'] . " is " . $row['age'] . "<br/>"                             
}
mysqli_close($connect);

?>

Answer 1

你吞咽了第一排，没有做任何事情;即改变：

$row= mysqli_fetch_array($result);

while ($row = mysqli_fetch_array($result))

为：

while ($row = mysqli_fetch_array($result))

Answer 2

您正在获取第一行，但没有对其执行任何操作，因此不会在while循环内获取此行，因此不会输出。

$row= mysqli_fetch_array($result); // remove this

while ($row = mysqli_fetch_array($result))
{
echo $row['first_name'] . " " . $row['last_name'] . " is " . $row['age'] . "<br/>"                             
}

Answer 3

在这部分：

$row= mysqli_fetch_array($result);

while ($row = mysqli_fetch_array($result))
{

你正在跳过第一行。你获取它，然后在while循环中你立即获取下一个。您应该删除第一次获取并仅使用：

while ($row = mysqli_fetch_array($result))
{

Answer 4

<?php
    $connect= mysqli_connect("localhost", "root", "", "friends_list")
    or die('error connecting with the database');

    $query= "SELECT * FROM people WHERE age=19";

    $result= mysqli_query($connect, $query)
    or die('error querying the database');

    while ($row = $result->fetch_assoc()){
        echo $row['first_name'] . " " . $row['last_name'] . " is " . $row['age'] . "<br/>"                             
    }
    mysqli_close($connect);
?>

$result->fetch_assoc()将获取与sql条件匹配的所有内容，并且您可以像在上面那样避免进一步的错误。您可以在这里fetch_assoc() function

了解更多信息

MySQL SELECT语句只返回一行

4 个答案: