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PHP - Mysql_query Syntax Error
Hoping someone can help me, ive spent the last hour looking up examples of mysql_query(Insert) and changing my query but im still gettin a syntax error and I cant understand where. Ive tried removing apostrophies and putting the longest variable string outside of the speach marks without any joy.
mysql_query("INSERT INTO $QA_table (SUB_CATEGORY, QUESTION, FROM, FROM_EMAIL, DISPLAY) VALUES ('$sub_cat','$message','$from','$from_email','2')") or die(mysql_error()); I get the following error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM, FROM_EMAIL, DISPLAY) VALUES ('Range','test question bla bla bla','Phil','phi' at line 1 Similar TutorialsIs there something wrong this? If so, I'm not seeing it...I keep getting this error: Quote ( ! ) Parse error: syntax error, unexpected T_STRING in C:\Program Files\wamp\www\register.php on line 1059 Here's my code for registering a member: Code: [Select] <?php mysql_query (INSERT INTO fans (id, username, email, password, country, region, gender, status, account, subscription, time_registered, date_registered, birthdate, name, website, age, activation_key) VALUES ('', '{$_SESSION["user"]["username"]}', '{$_SESSION["user"]["email"]}', '{$_SESSION["user"]["password"]}', '{$_SESSION["user"]["country"]}', '{$_SESSION["user"]["region"]}', '{$_SESSION["user"]["gender"]}', 'Offline', 'Inactive', 'Unsubscribed', '{$_SESSION["user"]["time"]}', '{$_SESSION["user"]["date"]}', '{$_SESSION["user"]["birthdate"]}', '{$_SESSION["user"]["name"]}', '{$_SESSION["user"]["website"]}', '{$_SESSION["user"]["age"]}', '{$_SESSION["user"]["activation_key"]}')) or die ("Could not register member"); ?> Help, please and thank you! hey I am still little new in php can you help me in this one func.php function confirm_query($result_set){ if(!$result_set){ die(mysql_error()); } function get_subjects_all() { global $connection ; $sub_query = "SELECT * FROM subjects ORDER BY position ASC"; $subject_query = mysql_query($sub_query,$connection); confirm_query($subject_query) ; return $subject_query; } function get_pages_all($subject_id) { global $connection; $query = "SELECT * FROM pages WHERE subject_id = '$subject_id' ORDER BY position ASC"; $page_set = mysql_query($query, $connection); confirm_query($page_set); return $page_set; } test.php require_once("config.php"); require_once("func.php"); $subject_query = get_subjects_all(); while($subject = mysql_fetch_array($subject_query)) { echo "<a href= \"test.php?subj=" . urlencode($subject["id"]) . "\">{$subject['menu_name']} <br />"; } $an = $subject["id"] ; $page_set = get_pages_all($an); while($page = mysql_fetch_array($page_set)) { echo "<a href = \"test.php?page=" . urldecode($page["id"]) . "\">{$page['menu_name']} </a><br />" ; // echo "<b> <a href= '1.php?page={$page['id']}' >{$page['menu_name']} </a></b> <br />"; } in this line WHERE subject_id = '$subject_id' the query don't work and gives me sql error i tried with this one ' ".$subject_id." ' but not worked but if i changed $subject_id to a number it work and read the data from the table thanks in advanced . Hey, I found some code online that references a bunch of sql commands in there own functions that can be called from a require_once(), All the ones I have tried work expect mysql_query when its called it returns this error: Quote Warning: mysql_result() expects parameter 1 to be resource, boolean given in But once i paste the mysql_query into the main script it works fines. code below is mine and the online script. Webcode: Code: [Select] $sql = "SELECT cat_id, cat_parent_id, cat_name, cat_description, cat_image FROM tbl_category WHERE cat_parent_id = $catId ORDER BY cat_name"; /* Don't know what getPagingQuery does and couldnt find any reference to it so i removed it and replaced with simple query in my own code */ $result = dbQuery(getPagingQuery($sql, $rowsPerPage)); My Code: Code: [Select] $sql = "SELECT cat_id, cat_parent_id, cat_name, cat_description, cat_image FROM tbl_category WHERE cat_parent_id = $catId ORDER BY cat_name"; $result = dbQuery($sql); Include file: Code: [Select] $dbHost = "localhost"; //SQL Server $dbUser = "root"; // Database username $dbPass = ""; // Database password $dbName = "lh_shop"; // Database name $sqlCon = mysql_connect($dbHost, $dbUser, $dbPass) or die ('MYSQL connection Failed. ' . mysql_error()); mysql_select_db($dbName) or die('Cannot select database. ' . mysql_error()); // Database query functions function dbQuery($sql) { return mysql_query($sql) or die('Query failed. ' . mysql_error()); } Thanks, I really hope someone has had this issue I have been looking at this code most of the morning and do not have a clue what is wrong with the code. I am hoping its not a stupid mistake, can someone please help me out? thank you
<title>Inputing Travel Detials</title>
<header>
<h1 align="center"> Adding Travel Detials </h1>
<body>
<p> <center><img src="cyberwarfareimage1.png" alt="Squadron logo" style="width:200px;height:200px" style="middle"></center>
<table border="1">
<tr>
<td><a href="index.php"> Home Page </a></td>
<td><a href="administratorhomepage.html">Administrator Home Page </a></td>
<td><a href="viewhomepage.html">View Home Page </a></td>
<td><a href="Inputhomepage.html">Input Home Page </a></td>
<td><a href="traveldetials.html">Enter More Travel Detials </a></td>
</table>
</p>
<?php
include "connection.php";
$Applicant_ID = $_POST["Applicant_ID"];
$Method_Of_Travel = $_POST["Method_Of_Travel"];
$Cost = $_POST["Cost"];
$ETA = $_POST["ETA"];
$Main_Gate_Advised = $_POST["Main_Gate_Advised"];
$query = ("UPDATE `int_board_applicant` SET `Method_Of_Travel`=`$Method_Of_Travel', `Cost`=`$Cost', `ETA`='$ETA', `Main_Gate_Advised`='$Main_Gate_Advised' WHERE `Applicant_ID`='$Applicant_ID'");
$result = mysqli_query($dbhandle, $query)
or die(mysqli_error($dbhandle));
if($result){ echo "Success!"; }
else{ echo "Error."; }
// successfully insert data into database, displays message "Successful".
if($query){
echo "Successful";
}
else {
echo "Data not Submitted";
}
//closing the connection
mysqli_close($dbhandle)
?>
Ok this is puzzleing. I am geting "Could not delete 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 '1' at line 1". but its is deleting the entry that needs to be removed. The "1" is the entry. Just not sure what is causing the error. I do have another delete php but I have put that on the back burning for the time being.
<?php $con = mysqli_connect("localhost","user","password","part_inventory"); // Check connection if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit(); } else { $result = mysqli_query($con, "SELECT * FROM amp20 "); $amp20ptid = $_POST['amp20ptid']; // escape variables for security $amp20ptid = mysqli_real_escape_string($con, $_POST['amp20ptid']); mysqli_query($con, "DELETE FROM amp20 WHERE amp20ptid = '$amp20ptid'"); if (!mysqli_query($con, $amp20ptid)); { die('Could not delete data: ' . mysqli_error($con)); } echo "Part has been deleted to the database!!!\n"; mysqli_close($con); } ?> Hi guys
I have this code below and all works fine when submitting this online application apart from when someone types either ' # & into one of the comment fields in which it throws up the error. Have tried various fixes from across the internet but no joy. Can anyone offer suggestions?
<?php
$con = mysql_connect("localhost:3306","root","password");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db('sfapp', $con);
$sql="INSERT INTO 'sfapp' ('surname_add','forename_add','dob_add','hometele_add','mobiletele_add','homeadd_add','siblings_add','schoolname_add','headname_add','schooladd_add','schooltele_add','schoolem_add','alevel_add','personstate_add','nameprovided_add','pe_add','se_add','PredGrade_Art','PredGrade_AScience','PredGrade_BusStudies','PredGrade_Electronics','PredGrade_EnglishLang','PredGrade_EnglishLit','PredGrade_French','PredGrade_German','PredGrade_Geog','PredGrade_Graphics','PredGrade_History','PredGrade_Maths','PredGrade_SepScience','PredGrade_ProductDesign','PredGrade_Spanish','PredGrade_Other','Gender_Male','Gender_Female','Sub_EnglishLit','Sub_Maths','Sub_FurtherMaths','Sub_Biology','Sub_Chemistry','Sub_Physics','Sub_French','Sub_German','Sub_Spanish','Sub_Geography','Sub_History','Sub_RE','Sub_FineArt','Sub_Business','Sub_Computing','Sub_GlobPersp','Sub_DramaAndTheatre','Sub_PE','Sub_Dance','Sub_Politics','Sub_Psychology','Sub_Sociology','readprospect_chk','Sib_Yes','Sib_No','Current_Student_Yes','Current_Student_No','I_Understand_chk','Current_Education_chk','Local_Care_chk','Staff_Cwhls_chk','Sub_Film')
VALUES
('$_POST[surname_add]','$_POST[forename_add]','$_POST[dob_add]','$_POST[hometele_add]','$_POST[mobiletele_add]','$_POST[homeadd_add]','$_POST[siblings_add]','$_POST[schoolname_add]','$_POST[headname_add]','$_POST[schooladd_add]','$_POST[schooltele_add]','$_POST[schoolem_add]','$_POST[alevel_add]','$_POST[personstate_add]','$_POST[nameprovided_add]','$_POST[pe_add]','$_POST[se_add]','$_POST[PredGrade_Art]','$_POST[PredGrade_AScience]','$_POST[PredGrade_BusStudies]','$_POST[PredGrade_Electronics]','$_POST[PredGrade_EnglishLang]','$_POST[PredGrade_EnglishLit]','$_POST[PredGrade_French]','$_POST[PredGrade_German]','$_POST[PredGrade_Geog]','$_POST[PredGrade_Graphics]','$_POST[PredGrade_History]','$_POST[PredGrade_Maths]','$_POST[PredGrade_SepScience]','$_POST[PredGrade_ProductDesign]','$_POST[PredGrade_Spanish]','$_POST[PredGrade_Other]','$_POST[Gender_Male]','$_POST[Gender_Female]','$_POST[Sub_EnglishLit]','$_POST[Sub_Maths]','$_POST[Sub_FurtherMaths]','$_POST[Sub_Biology]','$_POST[Sub_Chemistry]','$_POST[Sub_Physics]','$_POST[Sub_French]','$_POST[Sub_German]','$_POST[Sub_Spanish]','$_POST[Sub_Geography]','$_POST[Sub_History]','$_POST[Sub_RE]','$_POST[Sub_FineArt]','$_POST[Sub_Business]','$_POST[Sub_Computing]','$_POST[Sub_GlobPersp]','$_POST[Sub_DramaAndTheatre]','$_POST[Sub_PE]','$_POST[Sub_Dance]','$_POST[Sub_Politics]','$_POST[Sub_Psychology]','$_POST[Sub_Sociology]','$_POST[readprospect_chk]','$_POST[Sib_Yes]','$_POST[Sib_No]','$_POST[Current_Student_Yes]','$_POST[Current_Student_No]','$_POST[I_Understand_chk]','$_POST[Current_Education_chk]','$_POST[Local_Care_chk]','$_POST[Staff_Cwhls_chk]','$_POST[Sub_Film]')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
?>
<?php
//if "email" variable is filled out, send email
if (isset($_REQUEST['pe_add'])) {
//Email information
$admin_email = $_REQUEST['pe_add'];
$forename = $_REQUEST['forename_add'];
$email = "autoreply@testing.com";
$subject = "Application";
$desc =
"Dear $forename
Thank you for submitting your online application, we will be in touch shortly.
"
;
//send email
mail($admin_email, "$subject", "$desc", "From:" . $email);
//Email response
echo "Thank you for contacting us!";
}
//if "email" variable is not filled out, display the form
else {
?>
If you are seeing this, you need to go back and fill out the Personal Email section!
<?php
}
header("location:complete.php");
mysql_close($con)
?>
Thanks in advance.
Hello all,
Appreciate if you folks could pls. help me understand (and more importantly resolve) this very weird error:
Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42000]: Syntax error or access violation: 1064 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 'ASC, purchase_later_flag ASC, shopper1_buy_flag AS' at line 3' in /var/www/index.php:67 Stack trace: #0 /var/www/index.php(67): PDO->query('SELECT shoplist...') #1 {main} thrown in /var/www/index.php on line 67
Everything seems to work fine when/if I use the following SQL query (which can also be seen commented out in my code towards the end of this post) :
$sql = "SELECT shoplist.*, store_master.store_name, item_master.item_name FROM shoplist, store_master, item_master WHERE shoplist.store_id = store_master.store_id AND shoplist.item_id = item_master.item_id";However, the moment I change my query to the following, which essentially just includes/adds the ORDER BY clause, I receive the error quoted above: $sql = "SELECT shoplist.*, store_master.store_name, item_master.item_name FROM shoplist, store_master, item_master ORDER BY purchased_flag ASC, purchase_later_flag ASC, shopper1_buy_flag ASC, shopper2_buy_flag ASC, store_name ASC) WHERE shoplist.store_id = store_master.store_id AND shoplist.item_id = item_master.item_id";In googling for this error I came across posts that suggested using "ORDER BY FIND_IN_SET()" and "ORDER BY FIELD()"...both of which I tried with no success. Here's the portion of my code which seems to have a problem, and line # 67 is the 3rd from bottom (third last) statement in the code below: <?php
/*
$sql = "SELECT shoplist.*, store_master.store_name, item_master.item_name
FROM shoplist, store_master, item_master
WHERE shoplist.store_id = store_master.store_id
AND shoplist.item_id = item_master.item_id";
*/
$sql = "SELECT shoplist.*, store_master.store_name, item_master.item_name
FROM shoplist, store_master, item_master
ORDER BY FIND_IN_SET(purchased_flag ASC,
purchase_later_flag ASC,
shopper1_buy_flag ASC,
shopper2_buy_flag ASC,
store_name ASC)
WHERE shoplist.store_id = store_master.store_id
AND shoplist.item_id = item_master.item_id";
$result = $pdo->query($sql);
// foreach ($pdo->query($sql) as $row) {
foreach ($result as $row) {
echo '<tr>';
print '<td><span class="filler-checkbox"><input type="checkbox" name="IDnumber[]" value="' . $row["idnumber"] . '" /></span></td>';
Thanks
Parse error: syntax error, unexpected T_STRING in C:\xampp\htdocs\mywork\unique.php on line 15 <html> <head> <title> </title> </head> <body bgproperties="fixed"> <?php $dbhost = 'localhost'; $dbuser = 'root'; $dbpass = ''; $con = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'mywork'; mysql_select_db($dbname, $con); $sql=mysql_query(insert into users (regno,name,gender,date,month,year,emailid,cell,paddress,caddress,incometype,incomeamt,dad,fyes,dadocup,mom,myes,momocup,password) VALUES ('$_POST[regno]','$_POST[name]','$_POST[gender]','$_POST[date]','$_POST[month]','$_POST[year]','$_POST[emailid]','$_POST[cell]','$_POST[paddress]','$_POST[caddress]','$_POST[incometype]','$_POST[incomeamt]','$_POST[dad]','$_POST[fyes]','$_POST[dadocup]','$_POST[mom]','$_POST[myes]','$_POST[momocup]','$_POST[password]')"); $sql1=mysql_fetch_array($sql); $result = @mysql_query($SQl1); $result="SELECT * FROM users WHERE regno='$regno'"; while($row = mysql_fetch_array($result)) { //echo $row['regno']."regno<br>"; //echo $row['name']."name<br>"; //echo $row['gender']."gender<br>"; //echo $row['date']."date<br>"; //echo $row['month']."month<br>"; //echo $row['year']."year<br>"; //echo $row['emailid']."emailid<br>"; //echo $row['cell']."cell<br>"; //echo $row['paddress']."paddress<br>"; //echo $row['caddress']."caddress<br>"; //echo $row['incometype']."incometype<br>"; //echo $row['incomeamt']."incomeamt<br>"; //echo $row['dad']."dad<br>"; //echo $row['fyes']."fyes<br>"; //echo $row['dadocup']."dadocup<br>"; //echo $row['mom']."mom<br>"; //echo $row['myes']."myes<br>"; //echo $row['momocup']."momocup<br>"; //echo $row['password']."password<br>"; } echo "Thanks for Register!"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; mysql_close($con); ?> <form name="security" action="index.php" method="post"> <input type="submit" value="click here to login"> </form> </body> </html> The issue is there seems to be NO syntax error. 1. There is no relevant code before or after this line. 2. Yes, one would think a ! should be there as did I (I didn't write the code), however, even with the ! it still gives the same error. if (function_exists('gzcompress')) die(FUNCTION_NOT_FOUND); is giving me a syntax error, unexpected 'if', expecting 'function' or 'const' I am updating this code from PHP5.3 to PHP7.4 and I can't figure out what the syntax problem is since PHP allows this. I am using Eclipse PHP to do the conversion. Code: [Select] <?php mysql_connect ("-","-","-") or die ('Error'); mysql_select_db ("-"); $out = mysql_query("SELECT * FROM guestbook ORDER BY id DESC"); while($row = mysql_fetch_assoc($out); --and this one if that braces is deleted { ----this is where im getting the error $name = $row['name']; $email = $row['email']; $txt = $row['comment']; $msg = "Are you sure you want to delete"; /* @var $_REQUEST <type> */ if (isset($_REQUEST ["action"]) && $_REQUEST["action"] == "del") { $id = intval($_REQUEST['id']); mysql_query("DELETE FROM guestbook WHERE id=$id;"); echo "<action=index.php>"; } echo "<font face='verdana' size='1'>"; echo "<table border='0'> <tr><td>Name: ".$name."</td></tr>"." <tr><td>Email: ".$email."</td></tr> <tr><td colspan='2'>Comment:</td></tr> <tr><td colspan='2' width='500'><b>".$txt."</b></td></tr> <tr><td><a onclick=\"return confirm('.$msg.');\" href='index.php?action=del&id=".$row['id']."'><span class='red'>["."Delete"."]</span></a> </td></tr> </table><br />"; echo "<hr size='1' width='500' align='left'></font>"; } ?> Kindly help me please. When i delete ({) the error will become the ( i dont know what to do already. Thanks. I have been trying to get my files to upload onto a computer and I receive this message: Parse error: syntax error, unexpected T_STRING in /home/content/19/6550319/html/listing.php on line 27. Line 27 is how the php logs into my SQL. The problem is that I was able to log in before. I just made changes to the form by adding a dropdown menu and price and now it says it doesnt parse. Can anyone figure this out. I will include the code without the login information because the forum is public but I did put the words left out for you to see where I took out the passcodes. Code: [Select] <?php //This is the directory where images will be saved $target = "potofiles/"; $target = $target . basename( $_FILES['photo']['name']); //This gets all the other information from the form $price=$_POST['price']; $gig=$_POST['giga']; $yesg=$_POST['yesg']; $pic=($_FILES['photo']['name']); $pic2=($_FILES['phototwo']['name']); $pic3=($_FILES['photothree']['name']); $pic4=($_FILES['photofour']['name']); $description=$_POST['iPadDescription']; $condition=$_POST['condition']; $fname=$_POST['firstName']; $lname=$_POST['lastName']; $email=$_POST['email'] // Connects to your Database mysql_connect ("left out", "left out", "left out") or die(mysql_error()) ; mysql_select_db("left out") or die(mysql_error()) ; //Writes the information to the database mysql_query("INSERT INTO listing (price,giga,yesg,photo,phototwo,photothree,photofour,iPadDescription,condition,firstName,lastName,email) VALUES ('$price', '$gig', '$yesg', '$pic', '$pic2', '$pic3', '$pic4', '$description', '$condition', '$fname', '$lname', '$email')") ; //Writes the photo to the server if(move_uploaded_file($_FILES['photo']['tmp_name'], $target)) { //Tells you if its all ok echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory"; } else { //Gives and error if its not echo "Sorry, there was a problem uploading your file."; } echo date("m/d/y : H:i:s", time()) ?> I just enabled error reporting and I am not that familiar with it. I know I have an error some where around line 33. I know I am missing a bracket or a comma or some other syntax error I just cannot find where the error is. Below is my script. Thanks for any help. Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>Airline Survey</title> <meta http-equiv="content-type" content="text/html; charset=iso-8859-1" /> <meta name="author" content="Revised by abc1234"/> </head> <body> <?php $WaitTime = addslashes($_POST["wait_time"]); $Friendliness = addslashes($_POST["friendliness"]); $Space = addslashes($_POST["space"]); $Comfort = addslashes($_POST["comfort"]); $Cleanliness = addslashes($_POST["cleanliness"]); $Noise = addslashes($_POST["noise"]); if (empty($WaitTime) || empty($Friendliness) || empty($Space) || empty($Comfort) || empty($Cleanliness) || empty($Noise)) echo "<hr /><p>You must enter a value in each field. Click your browser's Back button to return to the form.</p><hr />"; else { $Entry = $WaitTime . "\n"; $Entry .= $Friendliness . "\n"; $Entry .= $Space . "\n"; $Entry .= $Comfort . "\n"; $Entry .= $Cleanliness . "\n"; $Entry .= $Noise . "\n"; $SurveyFile = fopen("survey.txt", "w") } if (flock($SurveyFile, LOCK_EX)) { if (fwrite($SurveyFile, $Entry) > 0) { echo "<p>The entry has been successfully added.</p>"; flock($SurveyFile, LOCK_UN; fclose($SurveyFile); else echo "<p>The entry could not be saved!</p>"; } else echo "<p>The entry could not be saved!</p>"; } ?d> <p><a href="AirlineSurvey.html">Return to Airline Survey</a></p> </body> </html> I don`t get it, waht is wrong?! Code: [Select] <?php require_once 'auth.php'; if (!isset($_SESSION['SESS_VERIFY'])) { header("location: access-denied.php"); exit(); } if ($_SESSION['lang'] == 'Ro') { // setare data romania date_default_timezone_set('Europe/Bucharest'); $today = getdate(); $zi = $today['mday']; $luna = $today['mon']; $lunastring = $today['month']; $an = $today['year']; $data = $zi.$luna.$an; $data = (string)$data; $ora = date('H:i:s'); $msg = array(); $err = array(); $luni = array ( 1=>'Ianuarie', 2=>'Februarie', 3=>'Martie', 4=>'Aprilie', 5=>'Mai', 6=>'Iunie', 7=>'Iulie', 8=>'August', 9=>'Septembrie', 10=>'Octobrie', 11=>'Noiembrie', 12=>'Decembrie'); // comun const SQL_ERR = 'SQL statement failed with error: '; const ADD_MODEL = 'ADAUGA UN MODEL NOU'; . .many constants.. . } elseif ($_SESSION['lang'] == 'It') {... Thank you!
Hello everyone,
1 <?php
7 // Create connection
10 // Check connection
14 $firstname = $conn->real_escape_string($_REQUEST['firstname']); 25 $sql2 = "INSERT INTO countries VALUES ('$country')"; 27 $sql3 = "INSERT INTO Contacts (firstname, lastname, address, city, country, phone, email) VALUES ('$firstname', '$lastname', '$address', $city, $country, '$phone_number','$email')";
29 SELECT * FROM cities;
if($conn->query($sql2) === true){
if($conn->query($sql3) === true){ Hi folks, I am a complete n00b at php and mysql. I am teaching myself from books and the WWW, but alas I am stuck... the error I get is: Parse error: syntax error, unexpected T_STRING in X:\xampp\htdocs\search.php on line 7 here is the code: <?php mysql_connect ("localhost", "user", "password") or die (mysql_error()); mysql_select_db ("it_homehelp_test") or die (mysql_error()); $term = $_POST['term']; $sql = $mysql_query(select * from it_homehelp_test where ClientName1 like '%term%'); <<<------this is line 7 while ($row = mysql_fetch_array($sql)){ echo 'Client Name:' .$row['ClientName1']; echo 'Address:' .$row['Address1']; echo 'Phone:' .$row['Tel1']; } ?> Any help you can offer would be great. I can also post the ".html" file that creates the search bar if it is needed. Thanks I am getting syntax error, unexpected T_CONSTANT_ENCAPSED_STRING on the date stamp line below but I can't see what is causing the problem. when I comment it out the problem goes away so I know I it is there. please help. $email_message = "Form details below.\n\n"; $email_message = date("m/d/Y")"\n"; |
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