我想避免为同一页面中的每个查询重新连接数据库，也就是想在mysqli_connect之后重用连接结果，这样我就可以加快结果 . 我在主机名前面尝试了“p：” . 但我仍然低于警告 . 当我连接时没有持久的“p：”并且连接每个查询时，我得到预期的结果 .
Warning: mysqli_close(): Couldn't fetch mysqli in C:\xampp\htdocs\TodayReport\tr1.php on line 99
Notice: Undefined variable: Result1Con in C:\xampp\htdocs\TodayReport\tr1.php on line 142
Warning: mysqli_close() expects parameter 1 to be mysqli, null given in C:\xampp\htdocs\TodayReport\tr1.php on line 142
我在下面的网页上提到
How to use mysqli persistent connection across different pages
Php Mysqli Persistent connection error
下面是我的脚本（我已经最小化以便能够快速阅读）
<?php
$webpage = "<!DOCTYPE html><html><head><title>Record Count</title><script src='sorttable.js'></script></head>";
$FirstProjectTitle = "Result1";
$SecondProjectTitle = "Result2";
//constructing HTML layout
Global $LoginDetailsCon;
Global $LoginDetailsResult;
$LoginDetailsCon=mysqli_connect("p:98.168.2.14","root","root@123","resultdbmain") or die("Failed to connect to resultdbmain: " . mysqli_connect_error());
$LoginDetailsQuery="SELECT * FROM login WHERE `IsWorking` = 1 AND `level` = 1";
$LoginDetailsResult=mysqli_query($LoginDetailsCon,$LoginDetailsQuery);
while ($row=mysqli_fetch_array($LoginDetailsResult,MYSQLI_ASSOC)) {
$Result1Con=mysqli_connect("98.168.2.14","root","root@123","resultdb1") or die("Failed to connect to resultdb1: " . mysqli_connect_error());
$result1Query="SELECT * FROM `resultdb1table1` WHERE `SID` = " . $row["ID"] ." AND `KeyDate` > '" . date("Y-m-d") . " 00:00:00' ";
$result1Result=mysqli_query($Result1Con,$result1Query);
$i=0;
$ik=0;
$SName = $row["Name"];
while ($row=mysqli_fetch_array($result1Result,MYSQLI_ASSOC)) {
$i++;
$ik = $ik + $row["KeyStroke"];
if ($i==1) {
$started=$row["KeyDate"];
}
$Lastkeyed=$row["KeyDate"];
}
if ($i > 0) {
//constructing HTML layout with results
}
}
$webpage .= "</table>";
mysqli_close($Result1Con);
//$LoginDetailsCon=mysqli_connect("98.168.2.14","root","root@123","resultdbmain");
//if (mysqli_connect_errno()) { echo "Failed to connect to resultdbmain: " . mysqli_connect_error(); }
//$LoginDetailsQuery="SELECT * FROM login WHERE `IsWorking` = 1 AND `level` = 1";
//$LoginDetailsResult=mysqli_query($LoginDetailsCon,$LoginDetailsQuery);
while ($row=mysqli_fetch_array($LoginDetailsResult,MYSQLI_ASSOC)) {
$Result2Con=mysqli_connect("98.168.2.14","root","root@123","resultdb2") or die("Failed to connect to resultdb2: " . mysqli_connect_error());
$result2Query="SELECT * FROM `resultdb2table1` WHERE `ID` = " . $row["ID"] ." AND `KeyDate` > '" . date("Y-m-d") . " 00:00:00' ";
$result2Result=mysqli_query($result2Con,$result2Query);
$i=0;
$ik=0;
$UName = $row["Name"];
while ($row=mysqli_fetch_array($result2Result,MYSQLI_ASSOC)) {
$i++;
$ik = $ik + $row["KeyStroke"];
if ($i==1) {
$started=$row["KeyDate"];
}
$Lastkeyed=$row["KeyDate"];
}
if ($i > 0) {
//constructing HTML layout with results
}
}
mysqli_close($result2Con);
mysqli_close($LoginDetailsCon);
$webpage .= "</body></html>";
echo $webpage ;
?>
1 回答
您对数据库连接的想法非常错误......不寻常 .
当然，你也不需要多个连接 . 只需使用一个连接即可 . 事实上，每次运行查询时都不必（也不应该）连接 . 相反，您应该为每个脚本执行连接 only once ，然后在代码中使用该单个连接变量 .
因此，创建一个名为
connection.php的文件，将连接代码放在那里一次，然后将include放在需要数据库交互的每个脚本中 .另外，请勿调用
mysqli_close(). PHP将为您关闭连接 .并暂时忘记持久连接 . 你不需要它 .