我用php创建了一个使用html代码的网页。我想将在网页中输入的用户信息保存到MySQL数据库中。我使用php作为中间人,将前端网页(htmnl代码)链接到数据库(mysql)。
在我的链接文件夹(中间人php文件)中,我有以下内容:
<?php
//Gets server connection credentials stored in serConCred2.php
require_once('ConCred2.php');
//SQL code for connection w/ error control
$con = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
if(!$con){
die('Could not connect: ' . mysqli_connect_error());
}
//Selection of the databse w/ error control
$db_selected = mysqli_select_db($con, DB_NAME);
if(!$db_selected){
die('Can not use ' . DB_NAME . ': ' . mysqli_error($con));
}
//Co-PI and Co-Investigator Information variables
$Co_FNAME = $_POST['fname'];
$Co_LNAME = $_POST['lname'];
$Co_SLNAME = $_POST['slname'];
$Co_DEGREE = $_POST['Degree_Selection'];
$Co_DEGREE_Other = $_POST['other_specify_degree']; //hold the value of degree if user selected other from the dropdown menu
$Co_CPOS = $_POST['Current_Position_Selection'];
$Co_CPOS_Other = $_POST['other_specify_cpos']; //hold the value of Current Position if user selected other from the dropdown menu
$Co_INST = $_POST['Institution_Selection'];
$Co_INST_Other = $_POST['other_specify_inst']; //hold the value of Current Position if user selected other from the dropdown menu
$Co_SCHOOL = $_POST['School_Selection'];
$Co_SCHOOL_Other = $_POST['other_specify_school']; //hold the value of Current Position if user selected other from the dropdown menu
$Co_DEPART = $_POST['Department_Selection']; //Este se estara eliminando en la version online
$Co_DEPART_Other = $_POST['other_specify_department']; //hold the value of Department if user selected other from the dropdown menu
$Co_PROGRAM = $_POST['program'];
$Co_EMAIL = $_POST['email'];
$Co_PHONE = $_POST['phone'];
//If decition when user select other from the dropdown menu
if($Co_DEGREE == "other_degree") $Co_DEGREE = $Co_DEGREE_Other;
if($Co_CPOS == "other_cpos") $Co_CPOS = $Co_CPOS_Other;
if($Co_INST == "other_inst") $Co_INST = $Co_INST_Other;
if($Co_SCHOOL == "other_school") $Co_SCHOOL = $Co_SCHOOL_Other;
if($Co_DEPART_Other == "other_department") $Co_DEPART = $Co_DEPART_Other;
//This sets a starting point in the rollback process in case of errors along the code
$success = true; //Flag to determine success of transaction
//start transaction
echo "<br>1. Going to set autocommit to 0";
$command = "SET AUTOCOMMIT = 0";
echo "<br>2. Autocomint has been set to 0";
echo "<br>3. Going to run query to see if result is true or false";
$result = mysqli_query($con, $command);
echo "<br>4. Finished running the query. Result is:" . $result;
echo "<br>5. Going to set command to BEGIN";
$command = "BEGIN";
echo "<br>6. Command is now BEGIN";
echo "<br>7. Going to run query for command BEGIN";
$result = mysqli_query($con, $command);
echo "<br>8. Query runned for command BEGIN";
echo "<br>9. Result value is: " . $result;
//Saves Pi values into database
/**
$sqlCoPI = "INSERT INTO co_pi_table (Fname, Lname, SLname, Degree, Current_Position, Institution, School, Department, Program, Email, Phone)
VALUES('$Co_FNAME', '$Co_LNAME', '$Co_SLNAME', '$Co_DEGREE', '$Co_CPOS', '$Co_INST', '$Co_SCHOOL', '$Co_DEPART', '$Co_PROGRAM', '$Co_EMAIL', '$Co_PHONE')";
*/
echo "<br>10. Going to write sql command to populate table pi_table";
/**
$sqlPi = "INSERT INTO pi_table (Fname, Lname, SLname, Degree, Current_Position, Institution, School, Department, Program, Email, Phone)
VALUES('$Co_FNAME', '$Co_LNAME', '$Co_SLNAME', '$Co_DEGREE', '$Co_CPOS', '$Co_INST', '$Co_SCHOOL', '$Co_DEPART', '$Co_PROGRAM', '$Co_EMAIL', '$Co_PHONE')";
*/
$sqlPi = "INSERT INTO pi_table (Fname) VALUES('$Co_FNAME')";
//Checks to see if theres an error in the pi db con
echo "<br>11. Sql command finished writting.";
echo "<br>12. Going to query the sql finished command to the database to determine value of result.";
$result = mysqli_query($con, $sqlPi);
echo "<br>13. Finished running sql command to database. Result value is: " . $result;
echo "<br>14. Going to enter if statements depending on result value";
if($result == false){
//die ('<br>Error in query to PI table: ' . mysqli_error($con));
echo "<br>15. I am inside the false statement. Success is going to be set as false. ";
$success = false;
//$success = true; //Cahnged this in order to test if values are being saved to db. Change back to false.
}
//Checks for errors or craches inside the code
// If found, execute rollback
echo "<br>16. Going to verify is success is true.";
if($success){
$command = "COMMIT";
$result = mysqli_query($con, $command);
//echo "<br>Tables have been saved with 0 errors.";
echo "<br><p style=\"color: red;\"Principal Investigator has been saved successfuly. <br><br>
You may now CLOSE this page and press the<br><br> \"Refresh List\" <br><br>
button to display name in dropdown menu selection.</p>";
}
else{
$command = "ROLLBACK";
$result = mysqli_query($con, $command);
echo "<br>17. Success was determined to be false.";
echo "<br>Error! Databases could not be saved.<br>
Contact system manager to report error. <br> <br>" . mysqli_error($con);
}
echo "<br>18. Setting autocommit back to 1 again.";
$command = "SET AUTOCOMMIT = 1"; //return to autocommit
$result = mysqli_query($con, $command);
//Displays message
//echo '<br>Connection Successfully. ';
//echo '<br>Database have been saved';
//Close the sql connection to dababase
mysqli_close($con)
?>
正如你所看到的,我要求用户填写他们的信息。所需的一些信息是下拉菜单字段,用户可以从这些字段中选择一个选项。
我遇到的问题是,当上面的php代码执行时,它确定$result变量为false并且不保存任何内容。执行代码时,将显示以下消息:
1. Going to set autocommit to 0
2. Autocomint has been set to 0
3. Going to run query to see if result is true or false
4. Finished running the query. Result is:1
5. Going to set command to BEGIN
6. Command is now BEGIN
7. Going to run query for command BEGIN
8. Query runned for command BEGIN
9. Result value is: 1
10. Going to write sql command to populate table pi_table
11. Sql command finished writting.
12. Going to query the sql finished command to the database to determine value of result.
13. Finished running sql command to database. Result value is:
14. Going to enter if statements depending on result value
15. I am inside the false statement. Success is going to be set as false.
16. Going to verify is success is true.
17. Success was determined to be false.
Error! Databases could not be saved.
Contact system manager to report error.
18. Setting autocommit back to 1 again.
为了安全起见,我不能发布html内容,因为它有敏感的名称信息和数据库。尽管我可以确保完全按照sql命令行中所述调用数据库中的表。
最佳答案
我找到问题了!
经过长时间的争论,我决定重新创建一个数据库,所有的信息都存储在其中。当我在sql命令中重定向表时(而不是将其保存在“pi_table”中,而是将其保存在一个名为“pi_table_2”的新创建的数据库中),一切都正常。
很快我的数据库就被破坏了,phpMyAdmin没有意识到它是被更新的。
作为参考,我的数据库表是InnoDB格式的。可能是什么导致了这种情况的发生,谁知道呢,但是如果你遇到了类似的问题,创建一个小的测试数据库并查看它是否保存。如果是,重新创建表,它可能会像解决我的问题一样解决您的问题。
再次感谢大家!!!!!