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php - 使用 PHP 和 MYSQL 显示 HTML

转载 作者:行者123 更新时间:2023-11-29 01:25:14 29 4
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用php和mysql显示html问题

嗨,基本上我有一个快速的 mysql_fetch_array 脚本来显示数据库中的内容,如下所示:

$sql_select_jobs = $db->query("SELECT * FROM `probid_jobs`");
while ($jobs_found = $db->fetch_array($sql_select_jobs)) {
$template_output .= "<div class=\"listed-jobs\" style=\"border: 1px solid gray; border-bottom: 0px; padding: 20px;\">";
$template_output .= "Job Title: " . $jobs_found['job_title'] . "<br />";
$template_output .= "Location: " . $jobs_found['location'] . "<br />";
$template_output .= "Salary: " . $jobs_found['salary'] . "<br />";
$template_output .= "Date Posted: " . $jobs_found['date'] . "<br /><br />";
$template_output .= "Description: " . $jobs_found['description'] . "...<a href=\"\">more</a><br />";

$template_output .= "</div>";
}

但是输出将如下所示:

Job Title: fff
Location: ff
Salary: fff
Date Posted: 18/06/10
Description: <b>fffffffffff <i>fffffffffffffffffff</i><br></b>...more

正如您所看到的,来自数据库的描述 html 没有被格式化为 html,由于某种原因它被转义了。

最佳答案

更改行:

$template_output .= "Description: " . $jobs_found['description'] . "...<a href=\"\">more</a><br />"; 

与:

$template_output .= "Description: " . html_entity_decode($jobs_found['description']) . "...<a href=\"\">more</a><br />"; 

更多信息:

http://www.php.net/manual/en/function.html-entity-decode.php

关于php - 使用 PHP 和 MYSQL 显示 HTML,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/3083303/

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