我有这个功能:
function findAllmessageSender(){
$all_from = mysql_query("SELECT DISTINCT `from_id` FROM chat");
$names = array();
while ($row = mysql_fetch_array($all_from)) {
$names[] = $row[0];
}
return($names);
}
返回私人消息系统中我的所有用户ID。然后我想得到user_id等于用户登录的所有消息,from_id等于我从上一个函数得到的所有from_id
:
function fetchAllMessages($user_id){
$from_id = array();
$from_id = findAllmessageSender();
$data = '\'' . implode('\', \'', $from_id) . '\'';
//if I echo out $ data I get these numbers '113', '141', '109', '111' and that's what I want
$q=array();
$q = mysql_query("SELECT * FROM chat WHERE `to_id` = '$user_id' AND `from_id` IN($data)") or die(mysql_error());
$try = mysql_fetch_assoc($q);
print_r($try);
}
print_r只返回1个结果:
Array (
[id] => 3505
[from_id] => 111
[to_id] => 109
[message] => how are you?
[sent] => 1343109753
[recd] => 1
[system_message] => no
)
但应该有4条消息。
答案 0 :(得分:4)
您必须为返回的每一行调用mysql_fetch_assoc()
。如果您只需拨打mysql_fetch_assoc()
一次,那么它只会返回第一行。
尝试这样的事情:
$result = mysql_query("SELECT * FROM chat WHERE `to_id` = '$user_id' AND `from_id` IN($data)") or die(mysql_error());
while ($row = mysql_fetch_assoc($result)) {
print_r($row);
}
答案 1 :(得分:0)
'mysql_fetch_assoc'返回一个与获取的行对应的关联数组,并向前移动内部数据指针。
您需要迭代数组:
while ($row = mysql_fetch_assoc($q)) {
echo $row["message"];
}