For the last 1 1/2 days I've been trying to store 16 row id's into a string and separate each id with a comma. The array I am getting is from MySQL. The error I am getting is
在过去的1 1/2天里,我一直试图将16行id存储到一个字符串中,并用逗号分隔每个id。我得到的数组来自MySQL。我得到的错误是
implode() function:passed invalid arguments
implode()函数:传递无效参数
$str=array();
$string="";
while($row = mysql_fetch_row($result))
{
$user_id=$row;
$str=$user_id;
foreach($str as $p=>$v){
comma($v);
}
}
function comma($v){
$string= implode(",",$v); echo $string;
}
3 个解决方案
#1
13
Try something like this:
尝试这样的事情:
$ids = array();
while ($row = mysql_fetch_assoc($result))
{
$ids[] = $row["UserID"];
}
echo implode(", ", $ids);
Replace "UserID"
with the columnname of the id in your table.
将“UserID”替换为表中id的列名。
So: first you build the array, next you implode the array into a string.
所以:首先构建数组,然后将数组内部转换为字符串。
#2
2
There is my solution:
有我的解决方案:
SELECT GROUP_CONCAT(UserID) as string FROM Users;
For this function the delimiter is ',' by default.
对于此函数,默认情况下分隔符为“,”。
#3
0
$query = 'SELECT id FROM your_table';
$rs = mysql_query($query);
$row = mysql_fetch_array($result);
return implode(',', $row);
the result 1,2,3...
结果1,2,3 ......
#1
13
Try something like this:
尝试这样的事情:
$ids = array();
while ($row = mysql_fetch_assoc($result))
{
$ids[] = $row["UserID"];
}
echo implode(", ", $ids);
Replace "UserID"
with the columnname of the id in your table.
将“UserID”替换为表中id的列名。
So: first you build the array, next you implode the array into a string.
所以:首先构建数组,然后将数组内部转换为字符串。
#2
2
There is my solution:
有我的解决方案:
SELECT GROUP_CONCAT(UserID) as string FROM Users;
For this function the delimiter is ',' by default.
对于此函数,默认情况下分隔符为“,”。
#3
0
$query = 'SELECT id FROM your_table';
$rs = mysql_query($query);
$row = mysql_fetch_array($result);
return implode(',', $row);
the result 1,2,3...
结果1,2,3 ......