Php 类mysqli的对象无法转换为字符串
我从以下代码中获取此错误:Php 类mysqli的对象无法转换为字符串,php,mysqli,runtime-error,Php,Mysqli,Runtime Error,我从以下代码中获取此错误: function updateRank($master_list, $event_id) { $cnt_ml = count($master_list); echo "count master list = $cnt_ml<br>"; $b=1; for ($k=0; $k<$cnt_ml; $k++) { echo "master list element 1 - ".$master_l
function updateRank($master_list, $event_id)
{
$cnt_ml = count($master_list);
echo "count master list = $cnt_ml<br>";
$b=1;
for ($k=0; $k<$cnt_ml; $k++)
{
echo "master list element 1 - ".$master_list[1]."<br>";
$foo = $master_list[$k];
// Update each team in event_team table
$update = "UPDATE event_team
SET pool_rank = $b
WHERE event_id = $event_id
AND team_id = $foo";
mysqli_query($conn, $update);// or die ('Could not run insert in event_team table');
echo "|".$update."|<br>"; // Leave this line for debugging the sql query.
$b++;
}
}
function updateRank($master\u list,$event\u id)
{
$cnt_ml=计数($master_list);
echo“计数主列表=$cnt_ml
”;
$b=1;
对于($k=0;$k变量$conn未声明。或者它是全局声明的,而您忘记添加
global $conn;
在updateRank函数中抛出了什么错误?^^^今年最好的评论您能给我们看一个数组的变量转储吗?错误消息是帖子的标题。抱歉。我应该说清楚。就是这样。“$conn”没有在全局声明。这解决了我的问题。非常感谢!!!!
global $conn;