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PHP - How To Check Before Insert With Xml
Hello PHP World,
I don't have access to the database yet for the project I am working on so I created a work around until I get full access to the mysql database. I Code: [Select] <?php $xml_file = "affiliate.xml"; // OPEN XML FILE ###################################################### if(!$xml=simplexml_load_file($xml_file)){ trigger_error('Error reading XML file',E_affiliate_ERROR); } if(!empty($_GET['add'])){ $u_name = stripslashes($_POST['name']); $u_company = stripslashes($_POST['company']); $u_address = stripslashes($_POST['address']); $u_city = stripslashes($_POST['city']); $u_state = stripslashes($_POST['state']); $u_phone = stripslashes($_POST['phone']); $u_email = stripslashes($_POST['email']); $u_desc = stripslashes($_POST['description']); $u_region = stripslashes($_POST['region']); // Generate new (appended) ID foreach($xml as $affiliate){ $last_id = $affiliate->uid; } $id = $last_id+1; // Add node $x = $xml->addChild("affiliate"); $x->addChild("uid",$id); $x->addChild("name",$u_name); $x->addChild("company",$u_company); $x->addChild("address",$u_address); $x->addChild("city",$u_city); $x->addChild("state",$u_state); $x->addChild("phone",$u_phone); $x->addChild("email",$u_email); $x->addChild("description",$u_desc); $x->addChild("region",$u_region); $xml->asXML($xml_file); } ?> html form Code: [Select] <br /> <br /> <h1>Affiliate </h1> <form name="xml_writing" method="post" action="?add=t"> <table width="393"> <tr><td width="56"> Name:<br /></td><td width="206"> <input type="text" name="name" width="150"/></td> <td>Region:</td> <td><select name="region" value="options"> <option value="resellerNE">NorthEast</option> <option value="resellerSE">SouthEast</option> <option value="resellerC">Central</option> <option value="resellerNE">NorthWest</option> <option value="resellerNE">SouthWest</option> <option value="resellerNE">International</option> </select></td> </tr> <tr> <td> Company:<br /></td><td> <input type="text" name="company" width="150" /></td> </tr> <tr><td> Address:<br /></td><td> <input type="text" name="address" width="150"/></td> </tr> <tr><td height="45"> City:<br /></td><td> <input type="text" name="city" width="150"/></td> <tr><td> State:<br /></td><td> <input type="text" name="state" width="150"/></td> <tr><td> Phone:<br /></td><td> <input type="text" name="phone" width="150"/></td> <tr><td> Email:<br /></td><td> <input type="text" name="email" width="150"/></td> </tr> <tr><td colspan="2"> Short description of yourself, your practice, your services, and other information of interest to potential clients or collaborators.<br /></td> <tr><td colspan="2"> <textarea name="description" style="width: 406px; height: 136px;"></textarea></td> </tr> <tr><td></td><td> <input type="submit" value="Submit" /></td> </tr> </table></form> What I am trying to do is have on submit check against the email node in the xml file to see if email exist. if exist string already a member else insert. if you can help with code or point to something that will help. "Everyone needs a mentor to lead them into success." Quote by Shelby Poston Similar TutorialsHello, i've got this code, to insert some form data into my db, it checks for a value if it exists already, it won't insert, now i want to add another value to check, but i don't know how. This is my code: Code: [Select] <?php if(isset($_POST['submit'])){ if (strlen($_POST['code']) == 12 && substr($_POST['code'],0,6) == '610147'){ $code = $_POST['code']; $select_query = mysql_query("SELECT * FROM jupiter WHERE code = '$code'"); if(mysql_num_rows($select_query) == 0){ $remote_addr = $_SERVER['REMOTE_ADDR']; $secret = $_POST['euro']; mysql_query("INSERT INTO jupiter(code,euro,ip,date,used) VALUES('$code','$secret','$remote_addr',CURDATE(),'n')"); } Print "<font color='green'>OK</font>"; } else { Print "<font color='red'>Error</font>"; } }?> This checks if "code" already exists, now i want to also have it check for "ip" Can someone help me out here please. Thanks ! I have displayed check box values(ugroup field) from ugroups table.now what i want to do is,when user select multiple check boxes and submit it should be insert into relavent feild in table.now it's insert check boxes values.but not in relevant field.this is my code.Please help me.
//select ugroup's from group table. <?php $result = "SELECT id,ugroup FROM group"; $res_result = db::getInstance()->query($result); ?>
<form action="db_sql/db_add_page.php" method="get">
Tittle :<input type="text" size="100" name="tittle" />
Description :<textarea cols="80" id="editor1" name="description" rows="10"></textarea>
//Display ugroups in textboxes and checkboxes
<?php
while( $line=$res_result->fetch(PDO::FETCH_ASSOC)) {
echo '<input type="checkbox" name="group[]" value=" '. $line['ugroup'] .'" />';
echo'<input type="text" name="ugroup" disabled="disabled" value=" '. $line['ugroup'] .'" size="7" "/>';
echo ' ';
}
?><input type="submit" value="Submit">
</form>
db_add_page.php
if(isset($_POST))
{
$tittle = $_POST['tittle'];
$description = $_POST['description'];
$ugroup = $_POST['group'];
$acc_status = "INSERT INTO add_services (id,tittle,description,g1,g2,g3,g4,g5,g6,g7,g8)
VALUES(NULL,'".$tittle."','".$description."','".$ugroup[0]."','".$ugroup[1]."','".$ugroup[2]."','
".$ugroup[3]."','".$ugroup[4]."','".$ugroup[5]."','".$ugroup[6]."','".$ugroup[7]."')";
$rate = db::getInstance()->exec($acc_status);
if(!$rate){
echo '<script type="text/javascript">alert("Update Error !");</script>';
}else{
header('Location:../add_page.php');
echo '<script type="text/javascript">alert("Successfuly Updated User Group !");</script>';
}
}
Attached Files
group.jpg 22.14KB
0 downloads
add_services.jpg 21.36KB
0 downloadsI have a register script, and I am wanting to make it so that if the username field contains, lets say "mod", "ass", and more, then it'll return an error and wont let them register. Hello, i am inserting some form data into my mysql db, i happen to get some duplicates so i want to check first if the entry exists already before i insert. my current code: Code: [Select] <?php if(isset($_POST['submit'])) { ?> <?php if (strlen($_POST['code']) == 19 && substr($_POST['code'],0,7) == '5541258') { mysql_query("INSERT INTO table(code,secret,ip,date) VALUES('$_POST[code]','$_POST[secret]','$_SERVER[REMOTE_ADDR]',CURDATE())"); Print "<font color='green'>The code will be checked now</font>"; } else { Print "<font color='red'>The code is invalid</font>"; } ?><?php } ?> I would like to use the value 'code' to check if the entry exists, that one is unique for each entry. How would i do that ? Thanks !
//database
create table mydata (
id int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,
fname varchar(20),
phoneno int(12) NOT NULL
/*......*/
);
//class my data php
<?php
include('connect.php');
class InsertMydata {
public function insertnow($fname, $phoneno) {
$connect = new Connect;
$insrt = $db -> prepare('INSERT INTO mydata (fname, phoneno) VALUES (?,?)');
$insrt -> execute(array($fname, $phoneno));
}
}
?>
//insernow validate form
<?php
include('../classs/mydata.php');
//Declare data and error arrays
$errors = [];
$mydara = [];
if(!preg_match('/^[a-zA-Z]{4,15}$/', $_POST['fname'])) {
$errors['fname'] = 'Enter full name!';
}
//this block not working even the phone exist
$connect = new Connect;
$phoneno = $_POST['phoneno'];
$checkiexist = $connect -> prepare('SELECT * FROM mydata WHERE phoneno = ?');
$checkiexist -> execute([$phoneno]);
if($checkiexist->rowCount() > 0) {
$errors['phonenoexist'] = 'Try another phone number!';
}
if(!empty($errors)){
$data['success'] = false;
$data['errors'] = $errors;
}else{
$data['success'] = true;
$data['message'] = 'success message!';
$mydata = new InsertMydata;
$mydata -> insertnow($fname, $phoneno);
}
echo json_encode($data);
?>
//my ajax
$("#insertbtn").click( function(e) {
var fname = $('#fname').val(),
phoneno = $('#phoneno').val(),
$.ajax({
url: 'insertnow.php',
type: 'POST',
data: {fname:fname, phoneno:phoneno},
dataType: "JSON",
encode: true,
}).done( function (data) {
if (data.success == false) {
if (data.errors.fname) {
$('#fname').append('<p class="text-danger">' + data.errors.fname + '</p>');
}
if (data.errors.phonenoexist) {
$('.card-header').append('<div class="alert alert-info alert-dismissible" role="alert">
<button type="button" class="close" data-dismiss="alert" aria-label="Close"><span aria-hidden="true">×</span></button>'+data.errors.phonenoexist+'</div>');
}
} else {
$('.card-header').append('<div class="alert alert-success alert-dismissible" role="alert"><button type="button" class="close" data-dismiss="alert" aria-label="Close"><span aria-hidden="true">×</span></button>'+data.message+'</div>');
}
});
e.preventDefault();
});
//the problem is, the code insert data even if the phone exist why?
the problem is, the code insert data even if the phone exist why? Edited April 8 by mahendaHi guys, im inserting data into the table using drop-down list & multi select list,well it works very well. but i need to make sure i should not insert same StudentID & CourseID twice. here my code for you could anyone tell me pls where should i write code to check existing data? <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("uni", $con)or trigger_error('MySQL error: ' . mysql_error()); $result = mysql_query("SELECT * FROM student") or trigger_error('MySQL error: ' . mysql_error()); echo '<select name="sid">'; while($row = mysql_fetch_array($result)) { echo '<option value="' . $row['StudentID'] . '">' . $row['StudentName'] . '</option>'; } echo '</select>'; // ---------------- ?> </div> <div class="style41" id="Layer7"> <?php $result = mysql_query("SELECT * FROM course") or trigger_error('MySQL error: ' . mysql_error()); echo '<select name ="cid[]" multiple="multiple" size="10">'; while($row = mysql_fetch_array($result)) { echo '<option value="' . $row['CourseID'] . '">' . $row['CourseName'] . '</option>'; } echo '</select>'; mysql_close($con); ?> ------------------------------------ <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("uni", $con)or trigger_error('MySQL error: ' . mysql_error()); if (!empty($_POST['sid']) && !empty($_POST['cid'])) { $ct = 0; $student = $_POST['sid']; foreach ($_POST['cid'] as $key => $course) { $sql = "INSERT INTO take (StudentID, CourseID) VALUES('".mysql_real_escape_string($student)."', '".mysql_real_escape_string($course)."')"; $query = mysql_query($sql) or trigger_error('MySQL error: ' . mysql_error()); if (mysql_affected_rows() > 0){$ct++;} } echo $ct . ' rows added.'; } mysql_close($con); ?> Hi, I need to insert some code into my current form code which will check to see if a username exist and if so will display an echo message. If it does not exist will post the form (assuming everything else is filled in correctly). I have tried some code in a few places but it doesn't work correctly as I get the username message exist no matter what. I think I am inserting the code into the wrong area, so need assistance as to how to incorporate the username check code. $sql="select * from Profile where username = '$username'; $result = mysql_query( $sql, $conn ) or die( "ERR: SQL 1" ); if(mysql_num_rows($result)!=0) { process form } else { echo "That username already exist!"; } the current code of the form <?PHP //session_start(); require_once "formvalidator.php"; $show_form=true; if (!isset($_POST['Submit'])) { $human_number1 = rand(1, 12); $human_number2 = rand(1, 38); $human_answer = $human_number1 + $human_number2; $_SESSION['check_answer'] = $human_answer; } if(isset($_POST['Submit'])) { if (!isset($_SESSION['check_answer'])) { echo "<p>Error: Answer session not set</p>"; } if($_POST['math'] != $_SESSION['check_answer']) { echo "<p>You did not pass the human check.</p>"; exit(); } $validator = new FormValidator(); $validator->addValidation("FirstName","req","Please fill in FirstName"); $validator->addValidation("LastName","req","Please fill in LastName"); $validator->addValidation("UserName","req","Please fill in UserName"); $validator->addValidation("Password","req","Please fill in a Password"); $validator->addValidation("Password2","req","Please re-enter your password"); $validator->addValidation("Password2","eqelmnt=Password","Your passwords do not match!"); $validator->addValidation("email","email","The input for Email should be a valid email value"); $validator->addValidation("email","req","Please fill in Email"); $validator->addValidation("Zip","req","Please fill in your Zip Code"); $validator->addValidation("Security","req","Please fill in your Security Question"); $validator->addValidation("Security2","req","Please fill in your Security Answer"); if($validator->ValidateForm()) { $con = mysql_connect("localhost","uname","pw") or die('Could not connect: ' . mysql_error()); mysql_select_db("beatthis_beatthis") or die(mysql_error()); $FirstName=mysql_real_escape_string($_POST['FirstName']); //This value has to be the same as in the HTML form file $LastName=mysql_real_escape_string($_POST['LastName']); //This value has to be the same as in the HTML form file $UserName=mysql_real_escape_string($_POST['UserName']); //This value has to be the same as in the HTML form file $Password= md5($_POST['Password']); //This value has to be the same as in the HTML form file $Password2= md5($_POST['Password2']); //This value has to be the same as in the HTML form file $email=mysql_real_escape_string($_POST['email']); //This value has to be the same as in the HTML form file $Zip=mysql_real_escape_string($_POST['Zip']); //This value has to be the same as in the HTML form file $Birthday=mysql_real_escape_string($_POST['Birthday']); //This value has to be the same as in the HTML form file $Security=mysql_real_escape_string($_POST['Security']); //This value has to be the same as in the HTML form file $Security2=mysql_real_escape_string($_POST['Security2']); //This value has to be the same as in the HTML form file $sql="INSERT INTO Profile (`FirstName`,`LastName`,`Username`,`Password`,`Password2`,`email`,`Zip`,`Birthday`,`Security`,`Security2`) VALUES ('$FirstName','$LastName','$UserName','$Password','$Password2','$email','$Zip','$Birthday','$Security','$Security2')"; //echo $sql; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } else{ mail('email@gmail.com','A profile has been submitted!',$FirstName.' has submitted their profile',$body); echo "<h3>Your profile information has been submitted successfully.</h3>"; } mysql_close($con); $show_form=false; } else { echo "<h3 class='ErrorTitle'>Validation Errors:</h3>"; $error_hash = $validator->GetErrors(); foreach($error_hash as $inpname => $inp_err) { echo "<p class='errors'>$inpname : $inp_err</p>\n"; } } } if(true == $show_form) { ?> I'm missing something here. I have a form, and when the submit is pressed, the relevant post data inserts into table one, then I want the last insert id to insert along with other form data into a second table. The first table's still inserting fine, but I can't get that second one to do anything. It leapfrogs over the query and doesn't give an error. EDIT: I forgot to add an error: I get: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'usage, why VALUES ('14', '', '123', '','1234', '', '')' at line 1 query:INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company) VALUES ('test 14', 'asdfa', 'asdf', 'adf','asdf', '', '', '', '123', '', '') Code: [Select] if (empty($errors)) { require_once ('dbconnectionfile.php'); $query = "INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company) VALUES ('$description12', '$sn', '$description4', '$cne','$description5', '$description6', '$description7', '$description8', '$description9', '$description10', '$description11')"; $result = @mysql_query ($query); if ($result) { $who_donated=mysql_insert_id(); $query2 = "INSERT INTO tbl_donation (donor_id, donor_expyear, donor_cvv, donor_cardtype, donor_authorization, amount, usage, why) VALUES ('$who_donated', '$donate2', '$donate3', '$donate4','$donate5', '$donate6', '$donate7')"; $result2 = @mysql_query ($query2); if ($result2) {echo "Info was added to both tables! yay!";} echo "table one filled. Table two was not."; echo $who_donated; //header ("Location: http://www.twigzy.com/add_plant.php?var1=$plant_id"); exit(); } else { echo 'system error. No donation added'; Can anyone tell me why this is not INSERTing? My array data is coming out just fine.. I've tried everything I can think of and cannot get anything to insert.. Ahhhh! <?php $query = "SELECT RegionID, City FROM geo_cities WHERE RegionID='135'"; $results = mysqli_query($cxn, $query); $row_cnt = mysqli_num_rows($results); echo $row_cnt . " Total Records in Query.<br /><br />"; if (mysqli_num_rows($results)) { while ($row = mysqli_fetch_array($results)) { $insert_city_query = "INSERT INTO all_illinois SET state_id=$row[RegionID], city_name=$row[City] WHERE id = null" or mysqli_error(); $insert = mysqli_query($cxn, $insert_city_query); if (!$insert) { echo "INSERT is NOT working!"; exit(); } echo $row['City'] . "<br />"; echo "<pre>"; echo print_r($row); echo "</pre>"; } //while ($rows = mysqli_fetch_array($results)) } //if (mysqli_num_rows($results)) else { echo "No results to get!"; } ?> Here is my all_illinois INSERT table structu CREATE TABLE IF NOT EXISTS `all_illinois` ( `state_id` varchar(255) NOT NULL, `city_name` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; Here is my source table geo_cities structu CREATE TABLE IF NOT EXISTS `1` ( `CityId` varchar(255) NOT NULL, `CountryID` varchar(255) NOT NULL, `RegionID` varchar(255) NOT NULL, `City` varchar(255) NOT NULL, `Latitude` varchar(255) NOT NULL, `Longitude` varchar(255) NOT NULL, `TimeZone` varchar(255) NOT NULL, `DmaId` varchar(255) NOT NULL, `Code` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; Hello, I'm having a bit of a problem here, all help to this issues would be much appreciated I am trying to use text boxes to insert numbers into the database based on what is inputed. If I have a string, like this for example: $variable = 09385493; And I want to insert it into the database like this: mysql_query("INSERT INTO integers(number) VALUES ('$variable')"); When checking the integers table in my database, looking at the number field, the $variable that was inserted is outputted as 9385493 Notice the number zero was taken out of the front of the number. If the number is double 0's (009385493), both of those zero's would disappear, too. Thanks Not being able to upload images in the directory with a path name in the database? 1) This is my code for insert into the database and directory:
<?php
$host="localhost";
$username="root";
$pass="";
$db="registration";
$conn=mysqli_connect($host,$username,$pass,$db);
if(!$conn){
die("Database connection error");
}
// insert query for register page
if(isset($_POST['ronel'])){
$images = $_FILES['file']['name'];
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["file"]["name"]);
// Select file type
$imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));
// Valid file extensions
$extensions_arr = array("jpg","jpeg","png","gif","pdf");
// Check extension
if( in_array($imageFileType,$extensions_arr) )
{
$details=$_POST['details'];
$location=$_POST['location'];
$checkbox=$_POST['checkbox'];
$injured=$_POST['injured'];
$agegender=$_POST['agegender'];
$contact=$_POST['contact'];
$empid=$_POST['empid'];
$dept=$_POST['dept'];
$organization=$_POST['organization'];
$summary=$_POST['summary'];
$name=$_POST['name'];
$outcome=$_POST['outcome'];
$cause=$_POST['cause'];
$action=$_POST['action'];
$reportedname=$_POST['reportedname'];
$position=$_POST['position'];
$organisation=$_POST['organisation'];
$reportedcontact=$_POST['reportedcontact'];
$reporteddept=$_POST['reporteddept'];
$status="Pending";
$comment=$_POST['comment'];
$query="INSERT INTO `proposals` (`details`,`location`,`date`,`time`,`checkbox`,`injured`,`agegender`,`contact`,`empid`,`dept` ,`organization`,`summary`,`image`,`outcome`,`cause`,`action`,`reportedname`,`position`,`organisation`,`reportedcontact`,`reporteddept`,`status`,`comment`) VALUES ('$details','$location', current_timestamp(),current_timestamp(),'$checkbox','$injured','$agegender','$contact','$empid','$dept' ,'$organization','$summary','$name','$outcome','$cause','$action','$reportedname','$position','$organisation','$reportedcontact','$reporteddept','$status','$comment')";
$res=mysqli_query($conn,$query);
if($res){
$_SESSION['success']="Not Inserted successfully!";
header('Location:');
}else{
echo "<script>alert('Proposal not applied!');</script>";
}
// Upload file
move_uploaded_file($_FILES['file']['tmp_name'],$target_dir.$image);
}
}
date_default_timezone_set("Asia/Kolkata");
?>
2) Here is the input file:
<form class="form-horizontal" method="post" action="" enctype="multipart/form-data">
<input type="hidden" name="ronel" value="">
<div class="form-group">
<label style="position:absolute; left:63%; top:425px;" for="inputEmail" class="col-lg-3"><b>Upload Images Here :</b></label><br><br>
<div class="col-lg-9">
<input style="position:absolute; left:78%; top:420px;" type="file" name="file" enctype="multipart/form-data" class="form-control" name="incident_reference" onchange="document.getElementById('inc_ref').src = window.URL.createObjectURL(this.files[0]); document.getElementById('inc_ref').className +='_active'; document.getElementById('inc_ref_span').className += '_hidden'">
</div><iframe id="inc_ref" class="form-group" width="220px" height="130px" style="position:absolute; left:78%; top:32%;"></iframe></div>
</form>
3) error code: Notice: Undefined index: name in /opt/lampp/htdocs/create-nearmiss.php on line 57 ONGC TRIPUR Edited March 28, 2020 by Ronel change of var Just a quick question I have a form, couple of drop down boxes, input fields etc.. at the end is the submit button. now my question is should my insert into database sqls go before the submit button inside the form? out side the form? after the button inside the form? If by any chance you could give me some advice on the best way to layout my sql that would be great. They are big and go into three tables, at the moment I have them as three seperate queries but I think they must be able to go into one. can someone please help me with the layout and syntax. here is what the queries look like at the moment: if ($IBselect=$_POST ['IBselect']); { //CUSTOMER $enterCust="INSERT INTO customer(username, password, title, firstName, lastName, address, town, country, postCode, phone, email, dateCust) VALUES ('$_POST[username]', '$_POST[password]', '$_POST[title]', '$_POST[firstName]', '$_POST[lastName]', '$_POST[address]', '$_POST[town]', '$_POST[country]', '$_POST[postCode]', '$_POST[phone]', '$_POST[email]', 'DATE: Auto CURDATE()', CURDATE()"; $enterCust_query=mysql_query($enterCust)or die(mysql_error()); //CARD $enterCard="INSERT INTO card(cardNumber, name, expDate, cardID) VALUES ('$_POST[cardNumber]','$_POST[name]', ' $_POST[expDate]', '$_POST[cardID]')"; $enterCard_query=mysql_query($enterCard); //BOOKING $RCenterBook=mysql_query("INSERT INTO booking (custNumber, cardID, rideName, seatNo1, seatNo2, price, price2, dateBook, ID_time_tbl) VALUES ('$custNumber', '$_POST[cardID]', '$rideName', '$_SESSION[IBextract]', '$_SESSION[IBextract2]', '$IBendPrice1', '$IBendPrice2', 'DATE: Auto CURDATE()', CURDATE()', '$_SESSION[IB_slot_Time]'"); $IBenterBooking_query=mysql_query($IBenterBook); if (!mysql_query($enterCard, $enterCust, $IBenterBook)) { die("Error:" .mysql_error()); } echo "1 record added IB"; } Thanks =) I get the following message: Quote name4Query failed: Unknown column 'late' in 'field list' with this code. What does it mean? Code: [Select] <?php $apt=$_POST['search_term']; $stat = mysql_connect("localhost","root",""); $stat = mysql_select_db("prerentdb"); $query = "SELECT name FROM payments WHERE late = 'L'"; $stat = @mysql_fetch_assoc(mysql_query($query)); echo $stat["name"]; $name=$_POST['name']; $apt=$_POST['apt']; $amtpaid=$_POST['amtpaid']; $rentdue=$_POST['rentdue']; $prevbal=$_POST['prevbal']; $hudpay=$_POST['hudpay']; $tentpay=$_POST['tentpay']; $datepaid=$_POST['datepaid']; $late=$_POST['late']; $comments=$_POST['comments']; $paidsum=$_POST['paidsum']; $query = " INSERT INTO payhist (name,apt,amtpaid,rentdue,prevbal, hudpay,tentpay,datepaid,late,comments,paidsum) VALUES('$name','$apt','$amtpaid','$rentdue','$prevbal', '$hudpay','$tentpay','$datepaid','$late','$comments','$paidsum')"; $stat = mysql_query($query) or die('Query failed: ' . mysql_error()); mysql_close(); echo "data inserted<br /><br />"; ?> Hello!Is this correct? Code: [Select] $sql="INSERT INTO pacienti (nume, prenume, cnp, varsta, sex, casa_asigurari, oras, strada, numar, judet, data_inregistrarii) VALUES ('$_POST[nume]','$_POST[prenume]','$_POST[CNP]','$_POST[varsta]','$_POST[sex]','$_POST[casa]','$_POST[oras]','$_POST[strada]','$_POST[nr]','$_POST[judet]','$_POST[internare]')"; $sql="INSERT INTO diagnostic (nume) VALUES ('$_POST[diagnostic]')"; and i mean..if i can do 2 insert in 2 different tables...or how can i do this? Would someone be able to tell me why I can't run a successful mysql_query($sql) with $sql="INSERT INTO myDB.Titles (Title, Year, cID) values('". $title . "', '" . $year . "', '" . $cID . "')"; $sql echoes out to "INSERT INTO myDB.Titles (Title, Year, cID) values('Test', '2000', '2')" It looks right but I'm probably off on the quotes somewhere. Thanks in advance. Dear Sir/Madame I am trying to insert comments into a table called comments using echo ID in the input form but so far i can insert the comments with the post id but unable re-comment again kindly help me what do i do?
This is the input form:
<div>
<form action="ehscomment.php" method="post">
<input type="hidden" name="id" value="<?php echo $id ?>">
<div>
<label>Add comment</label>
<div>
<textarea rows="6" cols="110" name="comment" placeholder="comment"></textarea>
</div>
</div>
<input type="submit" name="postcomment" value="comment"></form>
</div>
And this is ehscomment.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "registration";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST['postcomment'])){
$id = $_POST['id'];
$comment = $_POST['comment'];
$sql = "INSERT INTO comments (id,comment)
VALUES ('$id','$comment')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();}
?>
Error message: Error: INSERT INTO comments (id,comment) VALUES ('242','asddd') I'm trying to use PDO and get used to doing things this way. I've been away from php/mysql for a few years, so, I'm crusty. I'm not getting any error messages back on this code, but the insert just doesn't happen. My first guess is that I'm doing something wrong with the datetime now() function. But, I may not have the PDO code right. I tried the script the old fashion way with mysql_query() and that worked. So, it has to be something in this code. I believe my server is set up to do PDO as it shows: PDO PDO support enabled PDO drivers mysql, sqlite pdo_mysql PDO Driver for MySQL, client library version 5.0.45 My php version is 5.2.14 and Mysql is 5.0.45. Any help would be appreciated. Code: [Select] $DBH = new PDO("mysql:host=$host;dbname=$dbname", $user, $pass); $DBH->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION ); $sql=$DBH->prepare("INSERT INTO assets(asset_name,date_added,short_desc) VALUES (:asset_name,NOW(),:short_desc)"); $sql->bindParam(':asset_name',$asset_name); $sql->bindParam(':short_desc',$short_desc); $name=$_POST["input1"]; $short_desc=$_POST["input2"]; $DBH->exec(); echo $name; echo "\nPDO::errorInfo():\n"; print_r($DBH->errorInfo()); } catch(PDOException $e) { echo "Syntax Error: ".$e->getMessage(); } Okay, I'm hoping one of you can help me. I have a mysql database that I have configured through phpmyadmin. I have an android app that simply makes and sends a mysql query I can get it to successfully return values when using Select statements but when I use INSERT INTO, it returns " Error Query is invalid" BUT BUT BUT, when I use the same string and enter it through the sql tab in myphpadmin it works fine ! So here is the string ( the semicolons at the end of each field name are so I can use something common to split the string up when the data arrives back on the phone) Code: [Select] randomkey||||||INSERT INTO table4 (`geolat;` , `geolong;` , `mode;` , `destgeolat;` , `destgeolong;` , `cellphone;` , `email;` , `carrego;` , `colour;` , `rating;` , `comment;`) VALUES (0.0,0.0,'driver' ,-43.54779,172.62472, , '' ,'' , 'text' , 'ratingleftblank' , 'commentblank' ) the index4.php script is as follows Code: [Select] ?php /* * Written By: * James */ /************************************CONFIG****************************************/ //DATABSE DETAILS// $DB_ADDRESS="mysql1.openhost.net.nz"; $DB_USER="bling44"; $DB_PASS="sadlyinept"; $DB_NAME="bling44"; //SETTINGS// //This code is something you set in the APP so random people cant use it. $SQLKEY="randomkey"; /************************************CONFIG****************************************/ //these are just in case setting headers forcing it to always expire and the content type to JSON header('Cache-Control: no-cache, must-revalidate'); header('Content-type: application/json'); if(isset($_POST['tag'])){ //checks ifthe tag post is there $tag=$_POST['tag']; $data=explode("||||||",$tag); //split the SQL statement from the SQLKEY if($data[0]==$SQLKEY){ ///validate the SQL key $query=$data[1]; $link = mysql_connect($DB_ADDRESS,$DB_USER,$DB_PASS); //connect ot the MYSQL database mysql_select_db($DB_NAME,$link); //connect to the right DB if($link){ $result=mysql_query($query); //runs the posted query (NO PROTECTION FROM INJECTION HERE) if($result){ if (strlen(stristr($query,"SELECT"))>0) { //tests if its a select statemnet $outputdata=array(); while ($row = mysql_fetch_assoc($result)){ $outputdata[]=$row; //formats the result set to a valid array } echo json_encode(array("VALUE",$tag,array_merge($outputdata))); //sends out a JSON result with merged output data } else { echo json_encode(array("VALUE",$tag,array_merge(array(array("AFFECTED_ROWS ".mysql_affected_rows($link)))))); //if the query is anything but a SELECT it will return the array event count } } else echo json_encode(array("VALUE",$tag,array_merge(array(array("ERROR QUERY IS INVALID"))))); //errors if the query is bad mysql_close($link); //close the DB } else echo json_encode(array("VALUE",$tag,array_merge(array(array("ERROR Database Connection Failed"))))); //reports a DB connection failure } else { echo json_encode(array("VALUE",$tag,array_merge(array(array("ERROR BAD CODE SUPPLIED"))))); //reports if the code is bad } } ?> So to reiterate. I can search the DB but can't INSERT INTO, unless I go through the myphpadmin interface. Any ideas are very much appreciated I am wondering about the following problem: I have two sessions, one for the user ($_SESSION['user_id']) and second one for the products inside cart ($_SESSION['cart']). I need to insert data into table racun. I did it this way and it works, but i want to know if this can be done better? Code: [Select] while ($row=mysql_fetch_array($query)){ $id = $_SESSION['korisnik_id']; $idp = $row['product_id']; $kol = $_SESSION['cart'][$row['product_id']]['quantity'] ."<br>"; $insert = "INSERT INTO racun (product_id, quantity, korisnik_id) VALUES ('$idp', '$kol', '$id') "; $result = mysql_query($insert); } i am trying to update a column in the database. I can update int columns but what about text? i want to update it so that the new text is added to it while keeping the old one. Here is what i have(which isnt working): mysql_query("UPDATE ".DB_PREFIX."posts SET post_content = '$old_content' + '".$_POST['test']."' WHERE post_id = '$double_post_id'") or die(mysql_error()); i have also tried: SET post_content = post_content + '".$_POST['test']."' but that didnt work either. What should i be doing? |
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