php - 查询错误，试图通过关键字进行搜索

Question

我有一个变量和一个用户名，我想在用户名的字符串(函数描述)上搜索它

这有什么问题:

$function_keywords = mysql_real_escape_string($_POST['function_keywords']);
if($function_keywords=="" || empty($function_keywords)){
    redirect("show.php?functions=PHP");
}
//trim whitespace from the stored variable
$trimmed = trim($function_keywords);
//separate key-phrases into keywords
$trimmed_keywords = explode(" ",$trimmed);
// Build SQL Query for each keyword entered
foreach ($trimmed_keywords as $trimm){
// MySQL "MATCH" is used for full-text searching.
//this code is ebv weird , should check out soon!
$query = "SELECT * 
          FROM functions
          WHERE isEnabled=1 AND isPrivate=0
          AND function_description LIKE '{$trimm}' 
          AND user_name='{$user_name}'   
          ";
 // Execute the query to  get number of rows that contain search kewords
 $results=mysql_query ($query,$connection);

Answer 1

就“like”语法而言，您必须使用“%”符号。如果你查询

select * from table where column like '%yourkeyword%'

然后它会返回表列中包含“您的关键字”的所有行。

只有当 column = 'yourkeyword' 时，你的陈述才是正确的