php mysqli 查询结果显示所有行
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mysqli query results to show all rows
提问by Tom
I have the following code:
我有以下代码：
include $_SERVER['DOCUMENT_ROOT'].'/include/conn.php';
$query = "SELECT title FROM news_event";
$result = $mysqli->query($query);
$row = $result->fetch_array(MYSQLI_BOTH);
$row_cnt = $result->num_rows;
$result->free();
$mysqli->close();
This is fine if there is only one result as I can just echo $row['title']but if there are lots of results, how do I get this to loop through and print every row?
如果只有一个结果，这很好，因为我可以只回显，$row['title']但如果有很多结果，我如何让它循环并打印每一行？
I'm sure this is really simple but I'm just not sure what I need to search for in Google.
我确定这真的很简单，但我不确定我需要在 Google 中搜索什么。
I'm looking for a mysqliequivalent of this:
我正在寻找mysqli与此等效的内容：
while( $row = mysql_fetch_array($result) )
{
echo $row['FirstName'] . " " . $row['LastName'];
echo "<br />";
}
回答by raidenace
Just replace it with mysqli_fetch_arrayor mysqli_result::fetch_array:)
只需将其替换为mysqli_fetch_array或mysqli_result::fetch_array:)
while( $row = $result->fetch_array() )
{
echo $row['FirstName'] . " " . $row['LastName'];
echo "<br />";
}
Almost all mysql_*functions have a corresponding mysqli_*function.
几乎所有的mysql_*功能都有对应的mysqli_*功能。
回答by jaggedsoft
Simple mysqli solution:
简单的mysqli解决方案：
$db = new mysqli('localhost','user','password','database');
$resource = $db->query('SELECT * FROM table WHERE 1');
while ( $rows = $resource->fetch_assoc() ) {
print_r($rows);//echo "{$row['field']}";
}
$resource->free();
$db->close();
With Error Handling:If there is a fatal error the script will terminate with an error message.
带错误处理：如果出现致命错误，脚本将终止并显示错误消息。
// ini_set('display_errors',1); // Uncomment to show errors to the end user.
if ( $db->connect_errno ) die("Database Connection Failed: ".$db->connect_error);
$db = new mysqli('localhost','user','password','database');
$resource = $db->query('SELECT field FROM table WHERE 1');
if ( !$resource ) die('Database Error: '.$db->error);
while ( $row = $resource->fetch_assoc() ) {
echo "{$row['field']}";
}
$resource->free();
$db->close();
Using iterators:Support was added with PHP 5.4
使用迭代器：PHP 5.4 添加了支持
$db = new mysqli('localhost','user','password','database');
foreach ( $db->query('SELECT * FROM table') as $row ) {
print_r($row);//echo "{$row['field']}";
}
$db->close();
Fetch a single record:This code does not require a loop.
获取单个记录：此代码不需要循环。
$db = new mysqli('localhost','user','password','database');
$resource = $db->query('SELECT field FROM table');
$row = $resource->fetch_assoc();
echo "{$row['field']}";
$resource->free();
$db->close();
回答by Stegrex
Use:
用：
while ($row = $result->fetch_array(MYSQLI_BOTH)) {
// Look inside $row here, do what you want with it.
}
Look at the associative array examples here (should correspond with fetch_array() versions as well):
看看这里的关联数组示例（也应该与 fetch_array() 版本对应）：
回答by Dharman
You can simply loop using foreach
您可以简单地循环使用 foreach
$query = "SELECT title FROM news_event";
$result = $mysqli->query($query);
foreach($result as $row) {
echo $row['FirstName'] . " " . $row['LastName'];
echo "<br />";
}
If you would like to keep the mysqli code and HTML separate, then it's good idea to fetch all into an array and loop later.
如果您希望将 mysqli 代码和 HTML 分开，那么最好将所有内容提取到一个数组中并稍后循环。
$query = "SELECT title FROM news_event";
$result = $mysqli->query($query);
$news_events = $result->fetch_all(MYSQLI_ASSOC);
// Later in HTML:
<div>
<?php foreach($news_events as $row): ?>
<p><?= $row['FirstName'] ?> <?= $row['LastName'] ?></p>
<?php endforeach ?>
</div>
