使用PHP和MYSQL显示HTML

Question

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db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>";
} 

但是输出看起来像这样:

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

你可以看到db中的描述html没有被格式化为html,由于某种原因它被转义.

Answer 1

换行:

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

有:

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

更多信息:

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