dou9022 2013-12-15 03:08
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PHP - 内爆中的数组

I'm trying to put an array into a query but I doens't work. I'm tying it with implode() but then it gives me " Array to string conversion in ... on line 26". Why? With json_encode it worked out ...

Thanks for your help!

$sql = mysql_query("SELECT follows
            FROM follow 
             WHERE follower LIKE '".$id."'") or die (mysql_error());

if(mysql_num_rows($sql) < 1){
    echo "<br/>";
    echo "Follow someone";
} else {
    //Put all the id's of the users the user is following in an array. 
    $i = 0;
    $user_follows = array();
    while ( $row = mysql_fetch_assoc($sql) )
        {
            $user_follows[$i] = $row;
            $i++;
        }

    $user_follows = implode(" , ", $user_follows);
    echo $user_follows;
}
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1条回答 默认 最新

  • dongzuo9096 2013-12-15 03:13
    关注

    The second argument to implode must be an array of strings. But you're doing:

    $user_follows[$i] = $row;
    

    Since $row is an array, you're making an array of arrays (a 2-dimensional array), not an array of strings. That should be:

    $user_follows[] = $row['follows'];
    

    You don't need the $i variable, assigning to $array[] appends a new element to the array.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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