实际上，我连接了不同服务器中的3个数据库，但在mysqli_query中遇到了问题。
有时它工作得很流畅，没有任何错误，但有时它显示"PHP警告：mysqli_fetch_array（）期望参数1为mysqli_result"。。我不知道会发生什么，有时会发生。
这是我的联系方式：connection.php
<?php
$DatabaseName1 = "db_name1";
$DbHostName1 = "host1";
$DbUserName1 = "username";
$DbPassWord1 = "password";
$DatabaseName2q = "db_name2";
$DbHostName2 = "host2";
$DbUserName2 = "username2";
$DbPassWord2 = "password2";
$DatabaseName3q = "db_name3";
$DbHostName3 = "host3";
$DbUserName3 = "username3";
$DbPassWord3 = "password3";
$mysqli1 = mysqli_connect("$DbHostName1", "$DbUserName1", "$DbPassWord1", "$DatabaseName1");
$mysqli2 = mysqli_connect("$DbHostName2", "$DbUserName2", "$DbPassWord2", "$DatabaseName2q");
$mysqli3 = mysqli_connect("$DbHostName3", "$DbUserName3", "$DbPassWord3", "$DatabaseName3q");
$server[1] = $mysqli1;
$server[2] = $mysqli2;
$server[3] = $mysqli3;
$count_db = 3;
if(!$mysqli1){
echo "error to connect server 1st";
die();
}
if(!$mysqli2){
echo "error to connect server 2nd";
die();
}
if(!$mysqli3){
echo "error to connect server 3rd";
die();
}
if (mysqli_connect_errno()) {
printf("Connect failed: %s'n", mysqli_connect_error());
exit();
}
?>
task1.php
<?php
include("connection.php")
for($i=1;$i<=$count_db;$i++){
$conn = $server[$i];
$array = mysqli_fetch_array(mysqli_query($conn,"select * from `table`"));
$task = $array["field"];
}
?>
有人能帮忙吗？
喜欢这个吗
for($i=1;$i<=$count_db;$i++){
$conn = $server[$i];
$resultset = mysqli_query($conn,"select * from `table`")
if(!$resultset){
//do something
continue; //if you don't want to break it
}
$array = mysqli_fetch_array($resultset);
$task = $array["field"];
}