Question

我有这个功能：

function userPicBySkill($mainPassion){
$query  = mysql_query("SELECT id, username, imagename FROM users WHERE mainpassion =    '$mainPassion' ORDER BY RAND() LIMIT 5"); 

while ($row = mysql_fetch_array($query)) { 

$uid = $row["id"];
$username = $row["username"];   
$imagename = $row["imagename"];

echo "<a href='/$username'>  <img src='image/$imagename' width='40' height='41' alt =    '$username'></a>"; 
 } 
   }

如果我分配：

 $mainpassion = 'some skill';

我想做的是从我的表'技能'中随机选择5'技能'：

+----------+-----------+
|skill_id  | skill_name|
+----------+-----------+
|        1 |   guitar  |  
|        2 |   cooking |   
|        3 |   math    |    
|        4 |   plumbing|
|        5 |   piano   |
+----------+-----------+

然后为每项技能获得5张用户图片。我尝试了这个，但它没有显示任何内容：

function findRandomSkill(){

$skill_list = mysql_query("SELECT skill_name FROM skills ORDER BY RAND() LIMIT 5");
while($row = mysql_fetch_array($skill_list)){
$skill = $row['skill_name'];
echo '<div class="userBySkillDiv">
<h5>'.$skill.'</h5>';
userPicBySkill($skill);
echo'</div>';
}

Answer 1

最好应用联接而不是执行25次查询：

SELECT id, username, imagename 
FROM users 
INNER JOIN (SELECT skill_name
    FROM skills
    ORDER BY RAND() 
    LIMIT 5
) userskills ON users.mainpassion = userskills.skill_name
ORDER BY RAND()
LIMIT 5

Answer 2

你的意思是什么'没有显示任何东西'？是否返回了5个技能名单？

获得5项随机技能，试试$skill_list = mysql_query("SELECT DISTINCT(skill_name) FROM skills ORDER BY RAND() LIMIT 5");

Answer 3

在这样的函数中使用不同的$row变量，这会导致问题。

function userPicBySkill($mainPassion){
$query  = mysql_query("SELECT id, username, imagename FROM users WHERE mainpassion =    '$mainPassion' ORDER BY RAND() LIMIT 5"); 

while ($row1 = mysql_fetch_array($query)) { 

$uid = $row1["id"];
$username = $row1["username"];   
$imagename = $row1["imagename"];

echo "<a href='/$username'>  <img src='image/$imagename' width='40' height='41' alt =    '$username'></a>"; 
 } 
}

从sql中选择随机变量并将其分配给一个函数

3 个答案: