{if...else} 如果 mysql 字段等于 ex.1 如何显示图像

发布于 2024-12-08 00:30:39 字数 328 浏览 0 评论 0原文

我有一个问题：我有一个 mysql 表，其中包含诸如 pages_viewed 之类的值，每次用户访问页面（例如 1、2、3 等）时，该表都会递增。我的问题很简单，当字段等于或大于 1 时如何显示图像？

就像

select pages_viewed from table;
if field pages_viewed > 1 than echo images/image.png

原谅我的 php 一样，我不太擅长它。

我一直在这里和其他论坛上阅读，但我能找到的只是 drupal 或 wordpress。

原文

i have a question:
i have a mysql table that has values such as pages_viewed that increments every time a user accesses a page such as 1,2,3 etc.
My question is simple how do i display an image when the field equals or is more than 1?

select pages_viewed from table;
if field pages_viewed > 1 than echo images/image.png

pardon my php i'm not too good at it.

i've been reading here and in other forums but all i can find is drupal or wordpress.

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动次打次papapa 2024-12-15 00:30:39

类似的东西

header('Content-Type: image/png')
echo file_get_contents(images/image.png)

something like

header('Content-Type: image/png')
echo file_get_contents(images/image.png)

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無處可尋 2024-12-15 00:30:39

这里基本上就是除了数据库连接、表名、字段名之外的所有内容。

mysql_connect("localhost", "mysql_user", "mysql_password") or die("Could not connect: " . mysql_error());
mysql_select_db("mydb");

$result = mysql_query("select * from table");

while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
    # Any other code to post the page information
    if($row['pages_viewed'] > 0) echo "<img src='images/image.png' />";
    # Any other code to post the page information
}

Here is everything basically short of database connection, table name, field name.

mysql_connect("localhost", "mysql_user", "mysql_password") or die("Could not connect: " . mysql_error());
mysql_select_db("mydb");

$result = mysql_query("select * from table");

while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
    # Any other code to post the page information
    if($row['pages_viewed'] > 0) echo "<img src='images/image.png' />";
    # Any other code to post the page information
}

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