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PHP - Trouble With Insert Statement To Add Record
I am trying to create a query to insert data into a table in my Access database. I have the following query:
INSERT INTO Issues (DateRequested, CustomerID, ComputerID, Issue, ItemsIncl, ImageName) VALUES (#1/14/2015#, 1, 1, "Computer freezes while I'm on the internet.", "AC Adapter", "none.gif")which should be performed from within my PHP page. However, when I check the database afterward, the new record isn't there. I then tried performing the query directly in Access and it worked fine. Why would it work in Access, but not when I run the same query in PHP? Chris Similar TutorialsI have some code that will update a record and is generic, meaning any POST variables can be used - whatever you have on the form. See below: Code: [Select] $set = array(); foreach($_POST as $field => $value){ $field = mysql_real_escape_string($field); $value = mysql_real_escape_string($value); $set[] = "`{$field}` = '{$value}'"; } $query .= implode(", ",$set) . " WHERE $id_name = '".$id."' LIMIT 1"; mysql_query($query) or die(mysql_error()); My question is, how would I modify this to insert a NEW record (not update an existing one). I'm not sure how to order this within a foreach statement because the add query has a different form: insert into tabel (all the fieldnames here) VALUES (all the values here) <?php if(!isset($_SESSION)) { session_start(); } // UNCOMMENT NEXT LINE TO PRINT THE $_SESSION ARRAY TO THE SCREEN . . . // echo '<pre>'; print_r($_SESSION); echo '</PRE>'; if(empty($_SESSION['userID']) || $_SESSION['authorized'] != true ) { header("Location: login.php"); exit; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html lang="EN" dir="ltr" xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="content-type" content="text/xml; charset=utf-8" /> <meta http-equiv="refresh" content="5;url=editprofile.php"> <title>Saving Profile</title> </head> <body> <?php require_once ("dbconn.php"); ?> <?php $userID = $_SESSION['userID']; $insert_query = 'insert into users WHERE userID='$userID'( aim, msn, yim, psnID, xblGamertag, otherContact ) values ( "' . $_POST['aim'] . '", "' . $_POST['msn'] . '", "' . $_POST['yim'] . '", "' . $_POST['psnID'] . '", "' . $_POST['xblGamertag'] . '", "' . $_POST['otherContact'] . '" )'; mysql_query($insert_query); ?> Your profile has been saved! You will now be redirected from where you came from. <br /><a href="editprofile.php" title="Click here if you don't want to wait">Click here if you don't want to wait.</a> </body> </html> It creates a new record but I want it to update an existing one. It's an editprofile script. hi all ,, I have a table : test (id , name , title , area , city) and I want from user to insert more than one record .. I made the script but I don't know there are better than this code .. <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>My test</title> </head> <body> <table align="center" width="70%"> <tr> <form action="<?php $_SERVER['PHP_SELF']?>?>" method="get"> <?php if($_GET['submit']){ $con = mysql_connect("localhost","root",""); $db = mysql_select_db("test",$con); for($i = 1 ; $i < 10 ; $i++){ $name = $_GET['name'.$i]; $title = $_GET['title'.$i]; $area = $_GET['area'.$i]; $city = $_GET['city'.$i]; $sql = "INSERT INTO `test`.`test` (`title` ,`name` ,`area` ,`city`)VALUES ( '$name', '$title', '$area', '$city');"; $query = mysql_query($sql , $con); } }else { $con = mysql_connect("localhost","root",""); $db = mysql_select_db("test",$con) or die (mysql_error()); echo '<tr><td>Name</td><td>Title</td><td>Area</td><td>City</td></tr>'; for ($i = 1 ; $i < 10 ; $i++){ echo '<tr><td><input type="text" name="title'.$i.'" /></td> <td><input type="text" name="name'.$i.'" /></td> <td><input type="text" name="area'.$i.'" /></td> <td><input type="text" name="city'.$i.'" /></td></tr> <br /> '; } echo '<tr><td colspan="4"><input type="submit" name="submit" value="submit" /></td></tr>'; } ?> </form> </tr> </table> </body> </html> I wait you to advise me and I want your opinion .. than you very much .. Greetings mates, I have not been here for years. I have been thrust into an app written by someone else. The requirement is to amend, not edit, an existing record. In other words, rather than edit an existing record, management would like an empty new text box to enter amended data and save to another table so that when queried, would display existing record and the newly amended record. Below is an ID of an existing record:
$sql = "SELECT * FROM DBO.existingTable where year_created = 2019 Order by Id";
echo '<table id="results-table">
<thead>
<tr>
<th>ID</th>
<th>Edit</th>
<th>View PDF</th>
<th>Task Lead</th>
<th>Task Title</th>
<th>Division</th>
</tr>
</thead>';
while ( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_BOTH ) ) {
$div = ($row['division'] == "1" ? "Divname" : ($row['division'] == "2" ? "Highway" : "NULL"));
$taskType = ($row['task_type'] == 0 ? "Annual" : ($row['task_type'] == 1 ? "Carryover" : ($row['task_type'] == 2 ? "New" : "NULL")));
if($row['year_created'] == 2019){
echo '<tr id="taskId">
<td>'.$row['Id'].'</td>
<td><a href="project_edit.html?id='.$row['Id'].'" target="_blank">Edit</a></td>
<td><input type="button" onclick="getReport('.$row['Id'].')" value="View PDF"></td>
<td>'.$row['task_lead'].'</td>
<td>'.$row['task_title'].'</td>
<td>'.$div.'</td>
</tr>
}
}
echo '</table>'
The above code isa snippet of code in edit.php. What I would like to do is grab the ID which is '.$row['Id'].' from this edit page and use it when inserting the amended code which is in insert,php file:
//The ID from edit.php code should go here
$task_lead = (empty($_POST['task_lead'])) ? "''" : "'" . str_ireplace( "'", "", $_POST[ 'task_lead' ] ) . "'";
$task_title = (empty($_POST['task_title'])) ? "''" : "'" . str_ireplace( "'", "''", $_POST[ 'task_title' ] ). "'";
$division = (empty($_POST['division'])) ? "''" : "'" . str_ireplace( "'", "", $_POST[ 'division' ] ) . "'";
//Then my insert statement should be something like this:
$sql = "INSERT INTO DBO.Amended_Records ({id from table with existing record}, amended_taskActivities, amended_taskProducts,amendscope_dev_fhwa, amendconsultant_procurement, amendcontract_negotiations, amendconsultant_notice, amendtotal_duration, year_created)
VALUES ({$rowId - that I need to get from edit.php}, $amendproposed_activities, $amendanticipated_products,$amendscope_dev_fhwa,$amendconsultant_procurement,$amendcontract_negotiations,$amendconsultant_notice,$amendtotal_duration, 2021)";
When I insert this record into the DB, the id from edit.php is getting inserted as null. Can you please advise how I can resolve this? I have 2 tables, access_level and pages. In pages table i will have all the details of page like page_id, code, herf, pagename.. and i will select few from this table and store it in access_level table depending on the department and position. Now while editing, i will display all the pages which is stored in access_level pages with a particular page code along with those pages from pages table. if the page_id exists in access_level table, its has to get updated, if its not present then it should get inserted into access_level table. These things am doing with checkbox and <li>. my access_level table access_level.JPG 18.19KB 0 downloads and my pages table pages.JPG 37.86KB 0 downloads here is my code
$s1 = mysql_query("SELECT pages.page_id as pid, pages.code, pages.page,
pages.href, access_level.aid, access_level.page_id as pgid,
access_level.department, access_level.position,
access_level.active
FROM pages
LEFT JOIN access_level ON (pages.page_id=access_level.page_id
AND access_level.department=".$department."
AND access_level.position=".$position.") WHERE pages.code='snor die(mysql_error());
while($s2 = mysql_fetch_array($s1)) { ?>
<tr><td><li><?php echo $s2['page']; ?> </td><td><input type="checkbox" name="sn[]" value="<?php echo $s2['pid']; ?>" <?php if($s2['pgid'] === $s2['pid']) echo 'checked="checked"';?> />
<input type="hidden" value="<?php echo $s2['pid']; ?>" name="page_id[<?php echo $s2['pgid']; ?>]">
while submittings i am not getting the logic, how can be done. Please somebody suggest me
Not sure if this is right. I can't get it to insert my record. Can someone please tell me if I'm doing it right?
cus_functions.php
function dbRowInsert($table_name, $form_data)
{
global $conn;
$fields = array_keys($form_data);
$sql = "INSERT INTO ".$table_name."
(`".implode('`,`', $fields)."`)
VALUES('".implode("','", $form_data)."')";
return mysqli_query($sql);
}
process.php
include("new_db.php");
include("cus_functions.php");
$do=$_GET['do'];
if($do=='addpro'){
if(isset($_POST['submit'])){
$input = $_POST['title'];
$comp = '0';
$form_data = array('title' => $input, 'completed' => $comp);
dbRowInsert('projects', $form_data);
}}
I have a form that submits a record and saves it to the database, I've got that working already, I'm trying to figure out if there is a cleaner way to delete a record with a button that gets a value from the unique key, in this case 'id' here is how the delete button looks like: // get info from table $query = "SELECT id, Title, Message FROM table_name"; $result = mysql_query($query); //displaying all data while($row = mysql_fetch_assoc($result)) { echo" <div class='status'><h3>{$row['Title']} <br></h3>" . " <h5>{$row['Message']} <br><br></h5>"; $id = $row['id']; //trying to get unique key from database //delete button echo "</div> <form action='delete.php' method='post' /> <input type='hidden' name='delete' value='yes' /> <input type='hidden' name='id' value='$id' /> <input type='submit' value='remove' /></form>"; } As you can see, i'm trying to give $id a unique value but for some reason i'm not getting it. The delete.php code looks like this: if (isset($_POST['delete']))// check if delete was clicked { $query = "DELETE FROM table_name WHERE id='$id'"; echo "$id Deleted successfully"; } elseif(!mysql_query($query, $db_server)) { echo "DELETE failed: $query<br />" . myql_error() . "<br /><br />"; } mysql_close(); I'm new to php and still learning so, if you think there are other ways to do this, please let me know, the button won't do anything to the data. Thanks in advance I am trying to get this query correct. I want to insert a record into the database upon form submission but only if the record does not already exist. If the record exists, then I want it to be updated in the database.
What is happening: Upon form submit, a new record is entered into the database every time. Note: The contact_id column is both primary key and unique in my database. Here is my code:
if($_POST['submit']){
$con=mysqli_connect("localhost","username","password","database_name");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$org = mysql_real_escape_string($_POST['organization']);
$namefirst = mysql_real_escape_string($_POST['firstName']);
$namelast = mysql_real_escape_string($_POST['lastName']);
$emailaddy = mysql_real_escape_string($_POST['email']);
$phonenum = mysql_real_escape_string($_POST['phone']);
$appquestion = mysql_real_escape_string($_POST['appquestion']);
$banner = mysql_real_escape_string($_POST['banner']);
$bulletin = mysql_real_escape_string($_POST['bulletin']);
$giveaway = mysql_real_escape_string($_POST['giveaway']);
$app = mysql_real_escape_string($_POST['app']);
$tshirt = mysql_real_escape_string($_POST['tshirt']);
$tshirtp = mysql_real_escape_string($_POST['tshirtp']);
$print = mysql_real_escape_string($_POST['print']);
$party = mysql_real_escape_string($_POST['party']);
$orgnotes = mysql_real_escape_string($_POST['notes']);
$sql="INSERT INTO database_name (contact_id, first_name, last_name, email_address, phone_number, org, appquestion, banner, bulletin, giveaway, app, tshirt, promised_tee, print, party, org_notes)
VALUES
('','$namefirst','$namelast','$emailaddy','$phonenum','$churchorg','$appquestion','$banner','$bulletin','$giveaway','$app','$tshirt','$tshirtp','$print','$party','$orgnotes')
ON DUPLICATE KEY UPDATE first_name = '$namefirst', last_name = '$namelast', email_address = '$emailaddy', phone_number = '$phonenum', org = '$org', appquestion = '$appquestion', banner = '$banner', bulletin = '$bulletin', giveaway = '$giveaway', app = '$app', tshirt = '$tshirt', promised_tee = '$tshirtp', print = '$print', party = '$party', org_notes = '$orgnotes'" ;
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
}
From everything I have read, I need to use ON DUPLICATE KEY UPDATE to replace the old information with new information in the database upon form submission. While the insert part of my code is working, the portion with ON DUPLICATE KEY UPDATE is not working.
Why might this portion of the code not be working? Is there a better way to insert else update the information?
Thank you for any help or guidance you can give me! I've been working on this concept for three days and have read a ton of information about it, but am still not able to get it to work.
I'm trying to figure out why my entire if statement is not working properly. What is happening when I run my form is that it puts it into the first if statement regardless of what the value of $style is and I don't know why. The other parts of the for submission works BUT my if statement. And inside of firebug it is passing the RIGHT post data so it has the correct value for style each time. <?php // Include the database page require ('../inc/dbconfig.php'); if (isset($_POST['submitcharacter'])) { $charactername = mysqli_real_escape_string($dbc, $_POST['charactername']); $charactershortname = mysqli_real_escape_string($dbc, $_POST['charactershortname']); $sortorder = mysqli_real_escape_string($dbc, $_POST['sortorder']); $style = mysqli_real_escape_string($dbc, $_POST['style']); $status = mysqli_real_escape_string($dbc, $_POST['status']); $alignment = mysqli_real_escape_string($dbc, $_POST['alignment']); $division = mysqli_real_escape_string($dbc, $_POST['division']); $query = "INSERT INTO `characters` (charactername, charactershortname, status_id, style_id, division_id, alignment_id, sortorder, creator_id, datecreated) VALUES ('$charactername','$charactershortname','$status','$style','$division', '$alignment', '$sortorder', 1, NOW())"; mysqli_query($dbc, $query); $query_id = mysqli_insert_id($dbc); $query1 = "INSERT INTO `allies` (character_id) VALUES (".$query_id.")"; mysqli_query($dbc, $query1); $query2 = "INSERT INTO `rivals` (character_id) VALUES (".$query_id.")"; mysqli_query($dbc, $query2); if ($style = 1) { $query3 = "INSERT INTO `singles` (character_id) VALUES (".$query_id.")"; mysqli_query($dbc, $query3); } elseif ($style = 2) { $query4 = "INSERT INTO `tagteams` (character_id) VALUES (".$query_id.")"; mysqli_query($dbc, $query4); } elseif ($style = 3) { $query5 = "INSERT INTO `managers` (character_id) VALUES (".$query_id.")"; mysqli_query($dbc, $query5); } } elseif ($style = 4) { $query6 = "INSERT INTO `stables` (character_id) VALUES (".$query_id.")"; mysqli_query($dbc, $query6); } elseif ($style = 5) { $query7 = "INSERT INTO `referees` (character_id) VALUES (".$query_id.")"; mysqli_query($dbc, $query7); } else { $query8 = "INSERT INTO `staff` (character_id) VALUES (".$query_id.")"; mysqli_query($dbc, $query8); } ?> I have this as my code: if(!empty($stipulation)){ if($stipulation == "Championship Title Match"){ print "<h3 class=title>".$title." Match</h3>"; } else{ print "<h3 class=stipulation>".$stipulation."</h3>"; } } however I'm wanting to make it to where if it $stipulation has a value then it echos it and nothing if it doesn't have a value then don't nothing else happens. I also need a similiar if statement that will check to see if there is a value for $title and if there is then it echos $title Championship Match and again if no value exists it does nothing. I have the following code which seems to be in error in the conditional statement. The queries to the database produce the expected results, and the echo statements work OK. I want to make a selection based on whether data exists in the database for passState when passState=1. Can anyone tell me what is wrong with the conditional statement? Code: [Select] <?php $query1 = mysql_query("SELECT DISTINCT quizTitle, userId, passState, userScore, totalScore, DATE_FORMAT(userDate,'%b %e, %Y') AS userDate FROM quiz WHERE managerId = '$managerId' AND userId = '$userId' AND passState = 1 ORDER BY quizTitle ASC, passState DESC, userDate DESC "); $query2 = mysql_query("SELECT DISTINCT quizTitle, userId, passState, userScore, totalScore, DATE_FORMAT(userDate,'%b %e, %Y') AS userDate FROM quiz WHERE managerId = '$managerId' AND userId = '$userId' AND passState = 0 ORDER BY quizTitle ASC, passState DESC, userDate DESC "); ?> <table width="778" style="font-size:16px"> <tr> <?php if ($query1) { while ($row1 = mysql_fetch_array($query1)) { echo'<td width="13"></td> <td width="137" style="vertical-align:middle; text-align:left;">' ?> <?php echo "{$row1['quizTitle']} <br />\n"; ?> <?php echo '</td> <td width="13"></td> <td width="132" style="vertical-align:middle; text-align:left;;">' ?> <?php echo "{$row1['userDate']} <br />\n"; ?> <?php echo '</td> <td width="22"></td> <td width="112" style="text-align:center; text-align:center;">' ?> <?php if ("{$row1['passState']}" == 1) {echo "<img src=' ../wood/wood_tool_images/tick2.png' /><br />\n";}?> <?php echo '</td> <td width="30"></td> <td width="103" style="text-align:center; text-align:center;">' ?> <?php if ("{$row1['passState']}" == 0) {echo "<img src=' ../wood/wood_tool_images/cross2.png' /><br />\n";} ?> <?php echo '</td> <td width="19"></td> <td width="126" style="vertical-align:middle; text-align:center;">' ?> <?php echo "{$row1['userScore']}".'/'."{$row1['totalScore']} <br />\n"; ?> <?php echo '</td> <td width="23"></td> </tr>'; } }else{ while ($row2 = mysql_fetch_array($query2)) { echo '<td width="13"></td> <td width="137" style="vertical-align:middle; text-align:left;">' ?> <?php echo "{$row2['quizTitle']} <br />\n"; ?> <?php echo '</td> <td width="13"></td> <td width="132" style="vertical-align:middle; text-align:left;;">' ?> <?php echo "{$row2['userDate']} <br />\n"; ?> <?php echo '</td> <td width="22"></td> <td width="112" style="text-align:center; text-align:center;">' ?> <?php if ("{$row2['passState']}" == 1) {echo "<img src=' ../wood/wood_tool_images/tick2.png' /><br />\n";}?> <?php echo '</td> <td width="30"></td> <td width="103" style="text-align:center; text-align:center;">' ?> <?php if ("{$row2['passState']}" == 0) {echo "<img src=' ../wood/wood_tool_images/cross2.png' /><br />\n";} ?> <?php echo '</td> <td width="19"></td> <td width="126" style="vertical-align:middle; text-align:center;">' ?> <?php echo "{$row2['userScore']}".'/'."{$row2['totalScore']} <br />\n"; ?> <?php echo '</td> <td width="23"></td> </tr>'; } } ?> </table> Hey guys, I'm trying to get this working... No errors right now, but I'm not returning any results :/ been messing with it for days. Code: [Select] <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" /> <label for="search_by">Search By</label> <select name="search_by"> <option value"player">Player</option> <option value"city">City</option> <option value"alliance">Alliance</option> <option value"browse">Browse</option> </select> <input type="text" name"search"> <input type="submit" value="search" name="search"> <?php $search_by = $_POST['search_by']; $search = $_POST['search']; echo "<table><tr><td>Player</td><td>city</td><td>alliance</td><td>x</td><td>y</td><td>other</td><td>porters</td><td>conscripts</td><td>Spies</td><td>HBD</td><td>Minos</td><td>LBM</td><td>SSD</td><td>BD</td><td>AT</td><td>Giants</td><td>Mirrors</td><td>Fangs</td><td>ogres</td><td>banshee</td></tr>" ; $dbc = mysqli_connect('xx', 'xx', 'xx', 'xx') or die ('Error connecting to MySQL server'); $sql = "SELECT * FROM players WHERE ('$search_by') LIKE ('$search') "; //problem is here^^?? $result = mysqli_query($dbc,$sql) or die("Error: " .mysqli_error($dbc)); Not sure, any help would be greatly appreciated. Im trying to create a php template. What I want to do is take in GET data saying what page I am on, then include that in the template. THe trouble is I cant get the damn thing to work, it keeps saying the file does not exist! Here is my code <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <div class="content"> <?php $page = $_GET['page']; $path = '/inc ' .$page .".php"; if(!isset($page)) { die('No Page Specified'); } if (file_exists($path)) { include($path); } else { die('Does not exsist!'); } ?> </div> </body> </html> Having update statement trouble... The error I'm getting is... "Could not query the database - : " Not sure what I'm missing, I had this working before also, but now it's not. Here is my code for the update page... Code: [Select] <html> <head> <title>Update User</title> </head> <body> <form method="post" action="update_user2.php"> <?php $dbc = mysqli_connect('localhost', 'se266_user', 'pwd', 'se266') or die(mysql_error()); //delete users echo '<b>Delete or Update User</b>.<br />'; if (isset($_POST['remove'])) { foreach($_POST['delete'] as $delete_id) { $query = "DELETE FROM users WHERE course_id = $delete_id"; mysqli_query($dbc, $query) or die ('can\'t delete user'); } echo 'user has been deleted.<br />'; } if (isset($_POST['update'])) { $course_id = $_POST['course_id']; $course_name = $_POST['course_name']; $student_id = $_POST['student_id']; $query = "UPDATE users SET course_name ='". $course_name ."' WHERE course_id = $course_id"; $updres = mysqli_query($query); if(!$updres) { die(" Could not query the database - : <br/>". mysqli_error() ); } else { echo 'course has been updated.<br />'; } } //display users info with checkbox to delete $query = "SELECT * FROM users"; $result = mysqli_query($dbc, $query); while($row = mysqli_fetch_array($result)) { echo '<input type="checkbox" value="' .$row['course_id'] . '" name="delete[]" />'; echo ' ' .$row['course_name'] .' '. $row['student_id']; echo '<br />'; } mysqli_close($dbc); ?> <form method="POST" action="update_user2.php"> <label for="course_id">Course ID:</label> <input type="text" id="course_id" name="course_id" /> <br /> <label for="course_name">Course Name:</label> <input type="text" id="course_name" name="course_name" /> <br /> <label for="course_name">Student ID:</label> <input type="text" id="student_id" name="student_id" /> <br /> <input type="submit" name="remove" value="Remove" /> <input type="submit" name="update" value="Update" /> <br><br> </form> </body> </html> Can't figure out why my form is not deleting a user from database anymore. It was working a hour ago. Now it's not working, and I have no idea why. I've tried everything. Am I missing something? Here is the delete page... Code: [Select] <html> <head> <title>Update User</title> </head> <body> <?php $dbc = mysqli_connect('localhost', 'se266_user', 'pwd', 'se266') or die(mysql_error()); //delete users echo '<b>Delete or Update User</b>.<br />'; if (isset($_POST['remove'])) { foreach($_POST['delete'] as $delete_id) { $query = "DELETE FROM users WHERE course_id = $delete_id"; mysqli_query($dbc, $query) or die ('can\'t delete user'); } echo 'user has been deleted.<br />'; } if (isset($_POST['update'])) { foreach($_POST['update'] as $update_id) { $course_id = $_POST['course_id']; $course_name = $_POST['course_name']; $student_id = $_POST['student_id']; $query = "UPDATE `users` SET `course_name` = '$course_name' WHERE `course_id` = '$course_id' AND 'student_id' = '$student_id'"; mysqli_query($dbc, $query) or die ('can\'t update course'); $update_count = $db->exec($query); } echo 'course has been updated.<br />'; } //display users info with checkbox to delete $query = "SELECT * FROM users"; $result = mysqli_query($dbc, $query); while($row = mysqli_fetch_array($result)) { echo '<input type="checkbox" value="' .$row['course_id'] . '" name="delete[]" />'; echo ' ' .$row['course_name'] .' '. $row['student_id']; echo '<br />'; } mysqli_close($dbc); ?> <form method="POST" action="update_user2.php"> <label for="course_id">Course ID:</label> <input type="text" id="course_id" name="course_id" /><br /> <label for="course_name">Course Name:</label> <input type="text" id="course_name" name="course_name" /><br /> <label for="course_name">Student ID:</label> <input type="text" id="student_id" name="student_id" /><br /> </form> <form method="post" action="update_user2.php"> <input type="submit" name="remove" value="Remove" /> <input type="submit" name="update" value="Update" /> <br> <br> </body> </html> Hi all I'm working on a form, i've gotten all the values from the form to the process.php page and it is assigning the correct variables but i can't get it to insert into the database, i know the connection info is correct as i have other page connecting and i can create a recordset on that page. i guess my sql is wrong any help much appreciated. the following is the sql from the database and attached is the process.php page. -- Table structure for table `details` -- CREATE TABLE `details` ( `id` int(6) NOT NULL auto_increment, `first_name` varchar(20) NOT NULL, `sur_name` varchar(20) NOT NULL, `sex` varchar(6) NOT NULL, `age` varchar(3) NOT NULL, `house_no` varchar(5) NOT NULL, `street` varchar(150) NOT NULL, `town` varchar(50) NOT NULL, `bro_sis_cous_friend1` varchar(50) default NULL, `bscf_name1` varchar(150) default NULL, `and` varchar(4) NOT NULL, `bro_sis_cous_friend2` varchar(50) default NULL, `bscf_name2` varchar(150) default NULL, `his_her` varchar(4) default NULL, `him_her` varchar(4) default NULL, `from_name` varchar(150) NOT NULL, `photo_link` varchar(255) default NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=36 ; [attachment deleted by admin] could someone help me with this. here is what I have: <?php require_once('dbinfo.php'); ?> <?php mysql_connect($dbaddress,$username,$password); mysql_select_db($db) or die("Cannot find database!"); /* $query = "SELECT * FROM contacts"; $result = mysql_query($query); $numRows = mysql_numrows($result); $i = 0; while ($i < $numRows) { echo mysql_result($result, $i, "first") . " " . mysql_result($result, $i, "last"); $i++; } */ $fname = 'WrdScramble()'; $ftype = 'DAO'; $fdesc = 'Randomly rearranges the characters in a string or word.'; $fcode = ''; $query = "INSERT INTO functions VALUES ('','$fname','$fcode','$ftype','$fdesc')"; mysql_query($query); print($query); //$query = "SELECT fcode FROM contacts WHERE first = 'code'"; //$result = mysql_query($query); //$numRows = mysql_numrows($result); //$fcode = mysql_result($result, 0, "fcode"); ?> <? //close connection mysql_close(); ?> It is not inserting the data into the database, and if I use: mysql_error() and print it out, I get nothing with that either. I am not sure where to go! I also have used this: (fid, fname, fcode, ftype, fdesc) preceeding the 'VALUES' word in the statement just to see if it would take. No luck! Do I not see the obvious? Any help appreciated! thanks! (I have checked my field types and max length in PHPmyadmin and they are not an issue.) Hi, I would like to know how to write Insert data into a table statement where some of the data is coming from another table? I tried using Insert select statement but did not work. Here is what I am trying.... 1.........$sql = "INSERT INTO table1(id,fname, lname, gender, add1, add2, city, state, zip, country, email, phone, cellphone,employment,employmentinfo,dob, photo1,info) VALUES('$id','$fname','$lname','$gender','$add1','$add2','$city','$state','$zip','$country','$email','$phone','$cellphone','$employment','$employmentinfo','$dob','" . $image['name'] . "','$info') SELECT table2.id from table2 where table2.id=$id"; 2..........$sql = "INSERT INTO table1(id,fname, lname, gender, add1, add2, city, state, zip, country, email, phone, cellphone,employment,employmentinfo,dob, photo1,info) VALUES((SELECT id FROM table2 WHERE table2` WHERE username='".$_POST['username']."'),'$fname','$lname','$gender','$add1','$add2','$city','$state','$zip','$country','$email','$phone','$cellphone','$employment','$employmentinfo','$dob','" . $image['name'] . "','$info')"; I would appreciate your help. Thanks Smita I have this simple form that registers schools. This year I have decide to upgrade and include a feature that checks if the school is already registered or not based on the imputed name. Here is the code (that is the only way I know how): mysql_connect("", "", "") or die(mysql_error( '' )); mysql_select_db("") or die(mysql_error( '' )); $query = "SELECT * FROM School_Registrations WHERE School_Name= '$_POST[SchoolName]' "; $result = mysql_query($query); if (mysql_numrows($result) > 0) { while($row = mysql_fetch_array($result)) echo" error code here";} else {mysql_query("INSERT INTO `database`.`School_Registrations` (all the variables here);") or die(mysql_error( '' )); echo "Success Code";} I am trying to incorporate this code somewhere into the 'else' statement but I have no luck. I am constantly getting some errors and when I fix one there is one more to take its place. I am lost. The last one I can not fix and I am not sure what it wants from me: Fatal error: Call to undefined function getPage() It works by itself without the if/else statement but not in the code listed above $url = 'http://www.otherpage.com/page.php?'; $url .= 'email='.urlencode($_POST['email']); $result2 = getPage('', $url, '', 15); function getPage($proxy, $url, $header, $timeout) { $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_HEADER, $header); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt($ch, CURLOPT_PROXY, $proxy); curl_setopt($ch, CURLOPT_HTTPPROXYTUNNEL, 1); curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, $timeout); curl_setopt($ch, CURLOPT_REFERER, 'http://azsef.org'); curl_setopt($ch, CURLOPT_USERAGENT, 'Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.9.0.8) Gecko/2009032609 Firefox/3.0.8'); $result2['EXE'] = curl_exec($ch); $result2['INF'] = curl_getinfo($ch); $result2['ERR'] = curl_error($ch); curl_close($ch); return $result2; } Can you tell me why doesn't it work? Thanks I have a simple INSERT statement that isn't inserting anything into one of the columns: coin_name I've gone over everything and can't figure out why. All other columns get data. HTML form passes everything fine. Is there a good way to debug this? I've triple checked everything and there are no type-o's that I see. Driving me mad... $query = "INSERT INTO Coin (coin_name, coin_value, coin_condition, year_minted, face_value, purchase_price) VALUES ('$coin_name', '$coin_value', '$coin_condition', '$year_minted', '$face_value', '$purchase_price');"; |
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