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PHP - Xml To Mysql (insert Problem)
Similar TutorialsHey guys for some reason this code is not working i can't see a problem myself could someone please have a look and point the issue out to me. what i mean by it not working is it won't insert into the database or show a mysql_error. thanks in advance Code: [Select] $guestip = $_SERVER['REMOTE_ADDR']; $time = date('G:i'); $date = date("y-m-d"); $query = mysql_query("SELECT * FROM IP_Address") or die(mysql_error()); while($row = mysql_fetch_assoc($query)){ if($guestip != $row['ip']){ //insert into db. mysql_query("INSERT INTO IP_Address(id, ip, date, time) VALUES(NULL,'$questip','$date','$time')") or die(mysql_error()); echo "inserted in to database"; echo mysql_error(); }else{ // add hit count and update time and date. echo "already in db"; } } I have an old site written for PHP 5.4 and under and trying (very trying) to get it to work with PHP 7x without much luck. Due to all the changes in 7 my code is one big error message, but one thing at a time. I cannot get the follow code to work at all, even though it worked in PHP 5. Error:
QUERY ERROR: 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 'viewuser.php?u=666' id='member'>THE PREDATOR [666] was added to the hit' at line 1 I have tried at least 20+ different ways of doing this but just can't get the right syntax to get it inserted into MySQL, the code below is just the latest version. If I echo the a href line out, it works perfect. I am sure it is something ridiculously simple, but I have been 4 hours and counting on this now. Thanks
gangevent_add_2($gangdata['gangID'], "<a href='viewuser.php?u=".$r['userid']."' ".$csscode[$r['userlevel']-1].">".$r['username']."</a> [".$r['userid']."] was added to your hitlist");
function gangevent_add_2($gang, $text) {
global $db; $csscode;
$db->query("UPDATE users SET gangevent = gangevent + 1 WHERE gang={$gang}");
$db->query("INSERT INTO gangevents VALUES('','$gang', UNIX_TIMESTAMP(),'$text')");
}
I have problem with this code. It does absolutely nothing. When INSERT is over it should redirect to index.php but it does nothing. There is no error, when the submit is clicked the page just refresh itself and because of echo function it write all the values. What seems to be the problem (I going slightly mad ) Code: [Select] <?php require_once("public/includes/session.php"); ?> <?php require_once("public/includes/connection.php"); ?> <?php require_once("public/includes/functions.php"); ?> <?php include_once("public/includes/form_functions.php"); include_once("public/includes/header.php"); if (isset($_POST['submit'])) { // Form has been submitted. $errors = array(); $required_fields = array('nik', 'lozinka', 'ime', 'prezime', 'adresa', 'grad', 'postanskiBroj', 'fiskni', 'moblini', 'email'); $errors = array_merge($errors, check_required_fields($required_fields, $_POST)); $username = trim(mysql_prep($_POST['nik'])); $password = trim(mysql_prep($_POST['lozinka'])); $hashed_password = sha1($password); $ime = trim(mysql_prep($_POST['ime'])); $prezime = trim(mysql_prep($_POST['prezime'])); $adresa = trim(mysql_prep($_POST['adresa'])); $grad = trim(mysql_prep($_POST['grad'])); $postanskiBroj = trim(mysql_prep($_POST['postanskiBroj'])); $fiskni = trim(mysql_prep($_POST['fiksni'])); $moblini = trim(mysql_prep($_POST['mobilni'])); $email = trim(mysql_prep($_POST['email'])); echo $username . $hashed_password . $ime . $prezime . $adresa . $grad . $postanskiBroj . $fiskni . $moblini . $email; if ( empty($errors) ) { $query = " INSERT INTO `gume`.`korisnik` (`id`, `korisnicko_ime`, `lozinka`, `ime`, `prezime`, `adresa`, `grad`, `postanskiBroj`, `fiksni_telefon`, `mobilni_telefon`, `email`) VALUES (NULL, '$username', '$hashed_password', '$ime', '$prezime', '$adresa', '$grad', '$postanskiBroj', '$fiskni, $moblini', '$email' )"; $result = mysql_query($query, $connection) or die(mysql_error); if ($result) { redirect_to("index.php"); } else { $message = "The user could not be created."; $message .= "<br />" . mysql_error(); } } else { if (count($errors) == 1) { $message = "There was 1 error in the form."; } else { $message = "There were " . count($errors) . " errors in the form."; } } } else { // Form has not been submitted. $username = ""; $password = ""; } ?> <div id="telo"> <div id="kreiranjeNaloga"> <script src="SpryAssets/SpryValidationTextField.js" type="text/javascript"></script> <link href="SpryAssets/SpryValidationTextField.css" rel="stylesheet" type="text/css" /> <script src="SpryAssets/SpryValidationPassword.js" type="text/javascript"></script> <script src="SpryAssets/SpryValidationConfirm.js" type="text/javascript"></script> <link href="SpryAssets/SpryValidationPassword.css" rel="stylesheet" type="text/css" /> <link href="SpryAssets/SpryValidationConfirm.css" rel="stylesheet" type="text/css" /> <p>Polja sa * su obavezna</p> <form action="new_user.php" method="post"> <span id="sprytextfield1"> <label>Korisnicko ime: </label> <input type="text" name="nik" id="nik" size="40" value=""/> *<span class="textfieldMinCharsMsg">Korisnicko ime ne moze imati manje od 5 karaktera</span><span class="textfieldMaxCharsMsg">Korisnicko ime moze imati najvise 30 karaktera.</span></span><br /> <span id="sprypassword1"> <label>Lozinka:</label> <input type="password" name="lozinka" id="lozinka" size="40" value=""/> *<span class="passwordMinCharsMsg">Sifra mora sadrzati najmanje 5 karaktera.</span><span class="passwordMaxCharsMsg">Sifra moze imati najvise 30 karaktera.</span></span> <br /> <span id="spryconfirm1"> <label>Potvrdite lozinku:</label> <input type="password" name="password1" id="password1" size="40" value=""/> <span class="confirmRequiredMsg">*</span>Obe lozinke moraju da budu iste.</span> <br /> <span id="sprytextfield2"> <label>Ime:</label> <input type="text" name="ime" id="ime" size="40" value=""/> * </span> <br /> <span id="sprytextfield3"> <label>Prezima</label> <input type="text" name="prezime" id="prezime"size="40" value="" /> *</span> <br /> <span id="sprytextfield4"> <label>Adresa:</label> <input type="text" name="adresa" id="adresa" size="40" value=""/> * </span> <br /> <span id="sprytextfield7"> <label>Grad:</label> <input type="text" name="grad" id="grad" size="40" value="" /> * </span> <br /> </span><span id="sprytextfield9"> <label>Postanski Broj: </label> <input type="text" name="postanskiBroj" id="postanskiBroj" size="10" value=""/> * <span class="textfieldInvalidFormatMsg">Postanski broj nije pravilno upisan.</span></span><br /> <span id="sprytextfield5"> <label>Broj fiksnog telefona: </label> <input type="text" name="fiksni" id="Broj fiksnog telefona" size="40" value="" /> *<span class="textfieldInvalidFormatMsg">Broj telefona nije pravilno upisan</span></span> <br /> <span id="sprytextfield6"> <label>Broj mobilnog telefona: </label> <input type="text" name="mobilni" id="mobilni" size="40" value="" /> *<span class="textfieldInvalidFormatMsg">Broj telefona nije pravilno upisan</span></span><br /> <span id="sprytextfield10"> <label>Email:</label> <input type="text" name="email" id="email" size="40" value="" /> *<span class="textfieldInvalidFormatMsg">Email adresa nija pravilno upisana.</span></span><br /> <input name="submit" type="submit" id="submit" value="Kreiraj korisnika" /> </form> </div> </div> <?php include("public/includes/footer.php"); ?> i have created my own code of custom shopping cart
i have viewcart.php working great, now when i want to insert the orders from viewcart.php with list of like 5 items, how can i insert 5 names of products into my database 1 row
example
names are
1. jean
2. mond
3. richard
4. gwen
list above is the results of my while loop, now i want to insert those names to my database column[order_productname] so that i can identity what products are paid by my clients.
i tried fetch_array but if i assign variable to fetch array result, it only shows 1 which is "jean"
i wish this is possible
dforth
Code: [Select] $date = date('m-d-y'); $ip = $_SERVER['REMOTE_ADDR']; mysql_query("INSERT INTO users VALUES ($username, $password, 0, $ip, $date)") or die(mysql_error()); Error: 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 '.60.116, 03-06-11)' at line 1 I'm not sure why I get this error. :/ 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 Hi guys I have a registration form working fine, my database is as below: userid username password repeatpassword I have added another column which is "name", users can update their profile once they have logged in so I have created updateprofile.php and when I login-->go to update profile and insert my name nothing adds to mysql name column this is my code below: <?php include ("global.php"); //username session $_SESSION['username']=='$username'; $username=$_SESSION['username']; //welcome messaage echo "Welcome, " .$_SESSION['username']."!<p>"; if ($_POST['register']) { //get form data $name = addslashes(strip_tags($_POST['name'])); $update = mysql_query("INSERT INTO users (name) VALUES ('$_POST[name]') WHERE username='$username'"); } ?> <form action='updateprofile.php' method='POST'> Company Name:<br /> <input type='text' name='name'><p /> <input type='submit' name='register' value='Register'> </form> can you please tell me where in this code is wrong? Im new in php so please excuse me if I have silly mistakes. thanks in advance well this is truely embarrising...i have a insert statement which works within phpmyadmin but when using mysqli_query it returns a error.
INSERT INTO users (username, timestamp) VALUES ('test', UTC_TIMESTAMP())
Unknown column 'timestamp' in 'field list'
i've been playing about with this for a few hours now ...tried changing the column name (timestamp), adding ` around column names as well as table name.
the column exists which is the strangest part, and ive even checked there is no space after the column name in the db.
whats going on please?
I need help badly! What I want to do is insert into database the value from the selected radio group buttons.. All of them. There are 10 radio groups total (they can be less, but not more). Thanks! Code: [Select] <?php require_once('Connections/strana.php'); mysql_select_db($database_strana, $strana); ?> <link href="css/styles.css" rel="stylesheet" type="text/css" /> <table width="100%" height="100%" style="margin-left:auto;margin-right:auto;" border="0"> <tr> <td align="center"> <form action="" method="post" enctype="multipart/form-data" name="form1"> <table> <?php $tema = mysql_query("SELECT * from prasanja where tip=2")or die(mysql_error()); function odgovor1($string) { $string1 = explode("/", $string); echo $string1[0]; } function odgovor2($string) { $string1 = explode("/", $string); echo $string1[1]; } while ($row=mysql_fetch_array($tema)) { $id=$row['prasanje_id']; $prasanje=$row['prasanje_tekst']; $tekst=$row['odgovor']; ?> <tr> <td> </td> </tr> <tr> <td class="formaP"> <?php echo $prasanje?> </td> </tr> <tr> <td class="formaO"> <p> <label> <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor1($tekst) ?>" /> <?php odgovor1($tekst) ?></label> <br /> <label> <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor2($tekst) ?>" /> <?php odgovor2($tekst) ?></label> <br /> </p></td> </tr> <tr> <td> <br /> </td> </tr> <?php } ?> </table> <input align="left"type="submit" name="submit" value="Внеси" > </form> </td> </tr> </table> prasanje = question tekst/odgovor = answer The answer table: id - primary question_id - the questions ID whose answer is selected in the radio group user_id - cookie takes care of this answer - the value from radio group date - automatic I have this code: <?php $con = mysql_connect("localhost","hhh","hhh"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("hhh", $con); // -------------------- // Avatar insert check // -------------------- session_start(); $name = $_POST[name]; $group = $_POST[group]; $age = $_POST[age]; $usernameid = $_SESSION[id]; $result = mysql_query("SELECT * FROM avatars WHERE name='$_POST[name]'"); $num = mysql_numrows($result); if ($num == 0) { mysql_query("INSERT INTO avatars (id, usernameid, name, group, age, xp) VALUES ('', '$usernameid', '$name', '$group', '$age', '0')"); header( 'Location: me/' ) ; } else echo 'Sorry, please pick a new name'; ?> And it does everything but put the data into the datebase. If I add a session befor and after '$request' they both run, but the sql doesn't. No error returns, if just redirects to the other page. Any help? I don't understand where the empty value is. I've substituted the variables for text and still have the same problem. Code: Code: [Select] $sql = "INSERT INTO courses (course#, name, subject, semester, ap)VALUES('$courseNum', '$courseName', '$subject', '$semester', '$ap')"; Error: 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 '' at line 1 Hi,
The following code was written by someone else. It allows me to upload images to a directory while saving image name in the mysql table.
I also want the code to allow me save other data (surname, first name) along with the image name into the table, but my try is not working, only the images get uploaded.
What am I missing here?
if(isset($_POST['upload']))
{
$path=$path.$_FILES['file_upload']['name'];
if(move_uploaded_file($_FILES['file_upload']['tmp_name'],$path))
{
echo " ".basename($_FILES['file_upload']['name'])." has been uploaded<br/>";
echo '<img src="gallery/'.$_FILES['file_upload']['name'].'" width="48" height="48"/>';
$img=$_FILES['file_upload']['name'];
$query="insert into imgtables (fname,imgurl,date) values('$fname',STR_TO_DATE('$dateofbirth','%d-%m-%y'),'$img',now())";
if($sp->query($query)){
echo "<br/>Inserted to DB also";
}else{
echo "Error <br/>".$sp->error;
}
}
else
{
echo "There is an error,please retry or check path";
}
}
?>
joseph
Hi guys, I have an array: array ( [apple]=> red, [orange] => orange, [banana] => yellow ) That I want to insert into the apidata table, the PDO connection is good so I guess my code is wrong:
$mykeys = implode(', ', array_keys($newarr));
$myplaceholders[] = '(' . implode (", ", array_fill(0, count($newarr), '?')) . ')';
$values = array_values($newarr);
$res = $db->prepare("INSERT INTO apidata (item, value) VALUES $myplaceholders") ;
$res->execute([$mykeys, $values]);
The apidata table has three fields id (auto_increment) item and value, where am I going wrong? Thanks Hi all , sorry for posting question again , I really tried this for a whole day already but still can't get any solution... Now I completed my submit form with validation , when I click submit , above the form will echo the values entered by user and a button for confirmation , if everything okay then just click the confirm button , it will update into mysql . (I do this using php function ) But even if I click the confirm button , it will just refresh my page and won't update into mysql . I know that it is because this button has no related to the submit form so there is nothing for it to update , but I still don't know how to find other ways to do it . I even tried to put a disabled button first , when I click submit button then it will allow to click . Hmm...not a good way I know...(and won't works with my limited php knowledge) Can I get a little hints for this problem ? In my submit form Code: [Select] if(isset($_POST['Submit'])) { if($validator->ValidateForm()) { $myfunction->show_confirmation(); } For the function part: ( another php file ) Code: [Select] function show_confirmation(){ echo "Total recipient(s):".count($total)."<br>"; echo "<br>" ; echo "Recipient Number(s):<br>".$_POST['cellphonenumber']."<br>" ; echo "<br>" ; echo "Message:<br>".$_POST['inputtext']."<br>" ; echo "<br>" ; echo "Date:".$_POST['datetime']."<br>" ; echo "<br>"; echo "<br>"; echo " Proceed ?"; echo "<br>"; echo "<form>"; echo "<input type=\"submit\" name=\"Submit2\"> echo "</form>"; if(isset($_POST['Submit2'])) { $this->submit(); } } submit() is just a function that will insert values into mysql table . This is my normal form: After click submit button: So...what should I do to make the Submit2 button works ? Any hints or advices will be greatly appreciate . I'm not asking for a whole complete solution but just some hints...thanks for every reply . This is probably some obvious error I have made, but I cannot figure it out. I have made a few pages and now I am debugging them. My first page is called insert_purchase_order.php; on this page a person will enter some data in fields and hit the insert button. Then, the data is passed to another page, but when I try to insert into mysql it does not give me any errors, but I have no new rows either. The code for my 2nd page: Code: [Select] <?php session_start(); $action=$_GET[action]; if ($action==insert){ $randid=$_POST['randid']; $vendor=$_POST["vendor"]; $purchase_order_date=$_POST["purchase_order_date"]; $ship=$_POST["ship"]; $fob=$_POST["fob"]; $terms=$_POST["terms"]; $buyer=$_POST["buyer"]; $freight=$_POST["freight"]; $req_date=$_POST["req_date"]; $confirming_to=$_POST["confirming_to"]; $remarks=$_POST["remarks"]; $tax=$_POST["tax"]; $con = mysql_connect("localhost","root","pass"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("main", $con); mysql_query("INSERT INTO purchase_order (randid, vendor, purchase_order, ship, fob, terms, buyer, freight, req_date, confirming_to, remarks, tax) VALUES ($randid, $vendor,$purchase_order_date,$ship, $fob, $terms, $buyer, $freight, $req_date, $confirming_to, $remarks, $tax)"); mysql_close($con); echo 'Data Accepted...'; echo '<br/>'; echo 'P.O. Inserted Successfully'; }else{ echo 'Error... Please Contact Bruce.'; echo 'Bruce, no data was passed from the insert_purchase_order.php page.'; } ?> <a href="http://localhost/insert_purchase_order_items.php?po= <?php echo $randid; ?>">Insert Purchase Order Items</a> I have permissions and everything. Thanks Hi. I think you all know me by now so I'll cut to the chase. Code: [Select] <?php $host="edited"; $username="edited"; $password="edited"; $db_name="edited"; $tbl_name="topic"; // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $name=$_POST['name']; $detail=$_POST['details']; $sql="INSERT INTO $tbl_name(topic, detail, datetime)VALUES('$name', '$detail', NOW())"; $result=mysql_query($sql); if($result){ header("location:site.html");} else{ echo("I have failed you master.");} ?> Displayed error: "I have failed you master." Anyone know a possible cause? Thanks. Bye.
Array ( [data] => Array ( [0] => Array ( [latitude] => 22.934566 [longitude] => 79.08728 [type] => county [distance] => 44.328 [name] => Narsinghpur [number] => [postal_code] => [street] => [confidence] => 0.5 [region] => Madhya Pradesh [region_code] => MP [county] => Narsinghpur [locality] => [administrative_area] => [neighbourhood] => [country] => India [country_code] => IND [continent] => Asia [label] => Narsinghpur, India ) ) ) Hello Friends, Now i m stuck with simple insert query . here's the code Code: [Select] <?php include_once("conf.php"); $firstName=$_POST['fname']; $lastName=$_POST['lname']; $email=$_POST['email']; $dob=$_POST['dob']; $password=$_POST['pass']; $fname= stripslashes($firstName); $lname=stripslashes($lastName); $mail= mysql_real_escape_string($email); $password= mysql_real_escape_string($password); mysql_select_db('site'); $statement="Insert into Accounts(Name,lastName,emailId,DOB,password) VALUES($fname,$lname,$mail,$dob,$password)"; $query=mysql_real_escape_string($statement); mysql_query($query) or die("Cannot save data:</br> ".mysql_error()); echo "Data Saved Successfully"; ?> Now please explain me why i m getting the following error Cannot save data: 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 '@gmail.com,1988/12/20,password)' at line 1 Any help will be highly appreciated! |
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