Question

我一直在努力寻找答案。我想做的是（我认为）相当简单，但我找不到任何相关信息。

假设我的水果数据库中有一张表，我想做一些像......

<?php
require "../body-parts/config.php";

$result = mysql_query('SELECT * FROM fruits ORDER BY name');
$count = 1;
while ($fruit = mysql_fetch_array($result)) {
echo $count++ . " ";
echo $fruit['name'] . "<br>";
}
?>

现在让我们说输出是......

1 Apple

2哈密瓜

3葡萄

4 Kiwi

5柠檬

6 Orange

现在我想做的是获取Kiwi的数字（在这种情况下为4），以便我以后可以使用它...但我不知道我该怎么做。

简而言之，我如何获取返回值的数字位置？我不想回应它，我只想抓住它并将其粘贴在变量中，以便我可以再次使用它。

感谢任何帮助，谢谢。

Answer 1

$_kiwi_id = 0;


$result = mysql_query('SELECT * FROM fruits ORDER BY name');
$count = 1;
while ($fruit = mysql_fetch_array($result)) {
echo $count++ . " ";
echo $fruit['name'] . "<br>";
if(strtolower ($fruit['name']) == "kiwi")
$_kiwi_id = $count-1;
}

Answer 2

<?php
require "../body-parts/config.php";

$result = mysql_query('SELECT * FROM fruits ORDER BY name');

$storage = array();
while ($fruit = mysql_fetch_array($result)) {
$storage[] = $fruit['name'] . "<br>";
}

//later some point of time
echo $storage[3]['name'] ; //this will echo "kiwi" 
?>

Answer 3

使用数组：

<?php
require "../body-parts/config.php";

$result = mysql_query('SELECT * FROM fruits ORDER BY name');
$count = 1;
$fruits = array();
while ($fruit = mysql_fetch_array($result)) {
  $fruits[$count++] = $fruit['name'];
}

echo $fruits[4]; // prints Kiwi
?>

获取为某个值输出mysql_query的位置（编号）

3 个答案: