Question

我不久前继承了一个WordPress网站，该网站通过insert_php插件添加了一些自定义代码。它运行得很好，显然是我没有意识到的他们网站的重要组成部分。我不是程序员，只是所有行业IT人员的杰作。无论如何，我注意到最基本的MySQL命令可以用“ i”修复，并且mysqli_connect命令现在将接受dbname的第4个参数。我不确定在哪里截断代码它非常小，使用WordPress短代码（我将创建一个（真实的代码段））。我还阅读了查询，并且fetch_array遥遥无期。有什么建议吗？这是我的东西：

$con = mysql_connect("127.0.0.1","123xxxxxxx","4444444444444");
if (!$con) {
die("Can not connect:" . mysql_error());
}

mysql_select_db("dbxxxxx",$con);
$sql = 'SELECT * FROM EmployeeTable where PrimaryCampusPerson="Yes" order by SchoolDistrict, Site';

$myData = mysql_query($sql,$con);

echo "

";
while($record = mysql_fetch_array($myData)){
echo "

";
echo "";
echo "

";
echo "

";
echo "

";
}
echo "

Campus  District    Program Manager
" . "" . $record['Site'] . "" . "
" . $record['WorkAddress'] . "
" . $record['WorkCity'] . $record['WorkState'] . $record['WorkZip'] ."
" . $record['SchoolMainPhone'] ."   " . $record['SchoolDistrict'] . "   " . "" . $record['LastName'] . ", " . $record['FirstName'] . "" . "
" . $record['WorkCISPhone'] ."
" . "". $record['Email'] . "" ."
";

[/insert_php]````

Answer 1

 $con = mysqli_connect("127.0.0.1", "123xxxxxxx", "4444444444444", "dbxxxxx");
    // if there's an error connecting to db
 if (mysqli_connect_errno()) {
    echo "Failed to connect: " . mysqli_connect_errno();
 }

 $sql = "SELECT * FROM EmployeeTable
         WHERE PrimaryCampusPerson = "Yes"
         ORDER BY SchoolDistrict, Site";
 $myData = mysqli_query($con, $sql);

 while ($record = mysqli_fetch_array($myData)) {
    // echo db output
 }

Answer 2

$con = mysql_connect("127.0.0.1","123xxxxxxx","4444444444444");
mysql_select_db("dbxxxxx",$con);

以上两行可以合并为

$con=mysqli_connect("127.0.0.1","123xxxxxxx","4444444444444","dbxxxxx")

对于mysql查询mysql_query($sql,$con);

mysqli _query以connce作为第一个参数

mysqli_query($con,$sql);

对于mysql_fetch_array($myData)，您可以使用

mysqli_fetch_array($myData)

如何将此mysql旧代码转换为MYSQLI

2 个答案: