使用PHP在搜索中显示查询中的多行数据 [英] Display multiple rows of data from a query in a search using PHP

问题描述

一个需要的菜鸟确实是一个菜鸟！我将不胜感激任何帮助！



所以我连接到我的数据库，

我在MySQL中有一个名为testable的表，看起来像这样：



testtable



number（int）________ customer（string）

1________________________bill

2________________________韩语

3________________________bill

4________________________hank



我想搜索hank并创建一个包含数字的数组[2,4]

我所能做的就是让它找到第一行，然后它创建一个数组，即[2，'hank']



我尝试了什么：



我尝试制作一个php脚本的mysqli查询像这样：

$ sql =select * from testtable where customer ='hank';

$ result = mysqli_query（$ db，$ sql）;

$ row = mysqli_fetch_array（$ result，MYSQLI_NUM）;

echo（$ row [0]）;

echo（$ row [1]）;

？>

但它只返回：

hank2

我希望它返回

24



任何想法？

解决方案

sql =select * from testtable where customer ='hank';

< BLOCKQUOTE>结果= mysqli_query（

分贝，

a noob in need is a noob indeed! I would appreciate any help possible!

so I connect to my database,
I have a table named testable in MySQL that looks like this:

testtable

number(int) ________customer(string)
1________________________bill
2________________________hank
3________________________bill
4________________________hank

and I want to search for hank and make an array that contains the numbers [2,4]
All I can do is get it to find the first row and it makes an array that is [2,'hank']

What I have tried:

I tried making a php scripted mysqli query like this:
$sql="select*from testtable where customer='hank'";
$result=mysqli_query($db, $sql);
$row=mysqli_fetch_array($result, MYSQLI_NUM);
echo ($row[0]);
echo ($row[1]);
?>
but it only returns:
hank2
I want it to return
24

Any ideas?

解决方案

sql="select*from testtable where customer='hank'";

result=mysqli_query(

db,

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