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PHP - Using Variables As Part Of Insert Statement
I am new to PHP and am currently learning the ropes. I have writen some code to insert a line into a mySQL database. I have created 3 fields in the mySQL database and am passing values two them. I can use a declared variable to pass information to the filed ID (Index filed in mySQL) but if I use a variable to pass purely text values the database is not updated.
mysql_query( "INSERT INTO $tbl_name (category_Name, ID, test) VALUES ($category, $internalId, $category2)" ); If I replace the variables with specific text, the rows are added successfully. mysql_query( "INSERT INTO $tbl_name (category_Name, ID, test) VALUES ('werwerwer', $internalId, 'werrr')" ); It must be something stupid, I am sure. Full code below ob_start(); $host="localhost:8888"; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name="expenses"; // Database name $tbl_name="expense_category"; // Table name $category="test3"; $internalId="7"; $category2="test3"; // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); mysql_query( "INSERT INTO $tbl_name (category_Name, ID, test) VALUES ('werwerwer', $internalId, 'werrr')" ); echo "Done now 6!"; ?> Similar TutorialsHi, i have written a section of code for my website and i have been trying to get it to work but whatever i try it will not work Here is the code Code: [Select] $upgrade_user = mysql_query("SELECT * FROM user_info WHERE id='$id'"); while($row = mysql_fetch_array($upgrade_user)){ $real_balance = $row["$balance"]; $real_rank = $row["$rank"]; } echo $real_balance; if($real_rank == 'merchant' && $real_balance > '500'){ $n1x = '<a href="upgrades_info.php?userid=$id&rank=noble"><img src="images/noble.jpg" width="170" height="200" /></a>'; } else { $n1x = '<img src="images/noblex.jpg" width="170" height="200" />'; } It should outbut the first if, but instead it keeps displaying the else, i have checked the $real_rank and it matches merchant and the $real_balance is over 500. Any ideas? I am guessing its a simple error in the way i have written it, but i can't seem to get it to work. Thanks I am trying to build a news feed which shows the first 100 characters of news copy. I have it working to paste in the full text but am not sure how to insert only the first 100 characters. Code: [Select] <?php while($news_row=mysql_fetch_array($news)) { echo " <div class=\"news_box\"> <h3><a href=\"html/news.html\" rel=\"shadowbox; width=800; height=400;\">".$news_row['news_title']."</a> </h3> ".$news_row['news_copy']." </div>"; } ?> Can someone show me the correct code to display a group of checkbox values as part of an if statement? For example: in Programme One, lesson 100, lesson 310 and lesson 102 are available. In Programme Two, lesson 302, lesson 201, and lesson 102 are available. The "lesson types" are a checkbox group. I need to output the lesson types depending on whether they are part of Programme One or Two, along with other variables such as instructor student ratio. I put the following code together where field_lessons is the checkbox group; <?php if ($fieldsObjects['field_type']->data == "Programme One") { print '<p class="details_heading03">Instructor Student Ratio</p>'; echo $fieldsFormatted['field_student_ratio']. '<br />'; foreach ($fieldsObjects['field_lessons']->data as $name => $value) { echo $value; echo "<br/>"; } else { echo "no"; } ?> I have also tried using this statement: <?php if ($fieldsObjects['field_type']->data == "Programme One") { print '<p class="details_heading03">Instructor Student Ratio</p>'; echo $fieldsFormatted['field_student_ratio']. '<br />'; echo (isset($fieldsObjects['field_lessons']->data) && is_array($fieldsObjects['field_lessons']->data)) ? } else { echo "no"; } ?> I am new to php, and I am customising a template which is part of Sobi2, which has been installed as a component of Joomla! (CMS)? Thank you jo Hi, I would like to know how to write Insert data into a table statement where some of the data is coming from another table? I tried using Insert select statement but did not work. Here is what I am trying.... 1.........$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 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');"; Hello, guys. I am experiencing some problems with an INSERT statement in this page. It simply won't write to the database! I added echo at the bottom to check my variables and they print the values just fine. I checked the database, table and datafield names and everything is correct, plus I don't have any issues with the other 25 tables of my database. I'm using XAMPP btw... Any help would be appreciated! Code: [Select] <!DOCTYPE html> <html> <head> <meta content="text/html; charset=utf-8" http-equiv="content-type"> <title>Doctors</title> <link rel="stylesheet" href="style.css" media="screen" /> </head> <body > <?php session_start(); $inc_code=$_SESSION['incident']; $doc_code=$_REQUEST['doctor_code']; $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } $mdb = "registry_db"; mysql_select_db($mdb, $con); mysql_query("SET NAMES 'utf8'", $con); ?> <div id="myform"> <p> <h2>Doctor in charge</h2> </p> <?php $sql="INSERT INTO doctors_per_incident(Incident_code, doctor_code) VALUES ('$inc_code', '$doc_code')"; echo "1 record added"." ".$inc_code." ".$doc_code; mysql_close($con); ?> </div> </body> </html> I have a bunch of checkboxes I am pulling from a database, like so. Code: [Select] <?$true_query = mssql_query("SELECT * FROM checkbox_datbase ORDER BY ID ASC"); while ($true_row = mssql_fetch_assoc($true_query)) { $eth_id = $true_row['ID']; $eth_name = $true_row['Value']; echo "<input type=\"checkbox\" name=\"eth_$eth_id\" value=\"1\" class=\"input\" id=\"Ethnic_01\"/>$eth_name <br />"; } ?>How do I put these into a database that I have settup where it needs to match up with a userid. I am trying to do this, and can get it to look correctly by echoing out the various parts, but I can't put while loops into a variable, right? So how would I get all of this into one line in my $sql variable? Code: [Select] echo "INSERT INTO new_database (UserId,"; $true_query = mssql_query("SELECT * FROM checkbox_datbase ORDER BY ID ASC"); while ($true_row = mssql_fetch_assoc($true_query)) { $eth_id = $true_row['ID']; $eth_eth_id = $_POST['eth_' . $eth_id . '']; echo " $eth_id, "; } echo "DateUpdated) VALUES ('$user_ID', "; $true_query = mssql_query("SELECT * FROM checkbox_datbase ORDER BY ID ASC"); while ($true_row = mssql_fetch_assoc($true_query)) { $eth_id = $true_row['ID']; $eth_eth_id = $_POST['eth_' . $eth_id . '']; echo "'$eth_eth_id', "; } echo "'$currenttime')"; This echoes something that looks like this INSERT INTO user_ethnicity (UserId, 1, 2, 3, 4, 5, 6, 7, 8, DateUpdated) VALUES ('', '', '', '', '', '', '', '', '', '2012-01-23 13:24:18 PM') , which I need the actual statement to be $sql = "INSERT INTO user_ethnicity (UserId, 1, 2, 3, 4, 5, 6, 7, 8, DateUpdated) VALUES ('259', '', '', '1', '', '1', '', '', '', '2012-01-23 13:24:18 PM')"; Can anyone help me with this? 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 Folks, Tell me, do you see anything wrong in my INSERT ? If not, then why is it not INSERTING ? I get no php error, nor mysql error. Button I click. Then form data vanishes as if submitted. I check db and no submission came through!
<?php
//include('error_reporting.php');
ini_set('error_reporting','E_ALL');//Same as: error_reporting(E_ALL);
ini_set('display_errors','1');
ini_set('display_startup_errors','1');
?>
<form name = "submit_link" method = "POST" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<label for="domain">Domain:</label>
<input type="text" name="domain" id="domain" placeholder="Input Domain">
<br>
<label for="domain_email">Domain Email:</label>
<input type="email" name="domain_email" id="domain_email" placeholder="Input Domain Email">
<br>
<label for="url">Url:</label>
<input type="url" name="url" id="url" placeholder="Input Url">
<br>
<label for="link_anchor_text">Link Anchor Text:</label>
<input type="text" name="link_anchor_text" id="link_anchor_text" placeholder="Input Link Anchor Text">
<br>
<textarea rows="10" cols="20">Page Description</textarea>
<br>
<label for="keywords">Keywords:</label>
<input type="text" name="keywords" id="keywords" placeholder="Input Keywords related to Page">
<br>
<button type="submit">Click me</button>
<br>
<input type="reset">
<br>
</form>
<?php
if($_SERVER['REQUEST_METHOD'] === 'POST')
{
/*
if(ISSET($_POST['submit_link']))
{*/
mysqli_report(MYSQLI_REPORT_ALL|MYSQLI_REPORT_STRICT);
mysqli_connect("localhost","root","","test");
$conn->set_charset("utf8mb4");
if(mysqli_connect_error())
{
echo "Could not connect!" . mysqli_connect_error();
}
$query = "INSERT into links (domain,domain_email,url,link_anchor_text,page_description,keywords) VALUES (?,?,?,?,?,?)";
$stmt = mysqli_stmt_init($conn);
if(mysqli_stmt_prepare($stmt,$query))
{
mysqli_stmt_bind_param($stmt,'ssssss',$_POST['domain'],$_POST['domain_email'],$_POST['url'],$_POST['link_anchor_text'],$_POST['page_description'],$_POST['keywords']);
mysqli_stmt_execute($stmt);
mysqli_stmt_close($stmt);
mysqli_close($conn);
}
else
{
die("INSERT failed!");
}
//}
}
?>
@MacGuyver,
I will do the VALIDATIONS later. Remember, I was testing myself how much I can code without notes. What you see above was from my memory. Am still a beginner php student for 3yrs now, however! As for validation stuff will have to check notes and learn or relearn. Meaning, have to memorise the code before I add it here on current project. And so for now, ignore VALIDATIONS and let me know why it's not submitting data into db.
$price = 0.00; switch($someValue) { case "One": $price = 1; break; case "Two": $total = $price + 2; echo $total break; } I'm having problem on this. Basically what i want is to use the value of $price in case "One" to case "Two". But it won't work. Any idea how to accomplish this? Thank you in advance. This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=315503.0 I am trying to insert a record into a mysql database which has a $ as part of the value of a field. But it seems that when the INSERT query is processed its actually looking for a variable with that name and not just inserting the raw text. Here is the insert query $query = "INSERT mytable SET myfield='I would like to have this field contain this information with a $matches[1] showing in the field value as well'"; $matches[1] has no value. Its not a variable which is in this script. I am not trying to insert the value of it. I just want to insert the actual text characters $matches[1] How can I do that? I have no idea and there is no reason why this should not be working. im simply trying to add three variables into a database, and only one works. the other two do not work for any reason i can find. can someone point out my error, if any? code: <?php $date = date("Y-m-d"); $dbc = mysqli_connect('localhost', 'root', '', 'timer') or die('Error connecting to DB'); $query = @"INSERT INTO sessions (date, user, sessiontime) VALUES ('$date', '$user', '$sessiontime')"; $user = @$_GET['user']; $sessiontime = @$_GET['clock']; if (@$_GET['addDB'] == "Session Complete") { mysqli_query($dbc, $query) or die( '<br>Query string: ' . $query . '<br>Produced error: ' . mysqli_error($dbc) ); } ?> Form: Code: [Select] <label for="user"><b><em>Your name: </b></em></label><br /><input type="text" name="user" value="Admin/User" /> <input id="clock" name="clock" type="text" value="00:00:0" readonly><br> <input id="startstopbutton" type="button" value="S t a r t" onClick="startstop();" style="font-weight:bold"><br> <input type="submit" name="addDB" value="Session Complete" /> See, all the variables match up!? I dont get what im doing wrong? Hi, Im just having some trouble with this...maybe a fresh pair of eyes can help? Im getting a "Warning: mysqli_stmt::bind_param() [mysqli-stmt.bind-param]: Number of elements in type definition string doesn't match number of bind variables" error when I try run this: Code: [Select] $date = date("Y-m-d"); $header = $_POST['header']; $summary = $_POST['summary']; $content = $_POST['content']; $query = "INSERT INTO articles (pubdate, title, summary, content) VALUES(?, ?, ?, ?)"; $stmt = $mysqli->stmt_init(); if ($stmt->prepare($query)){ $stmt->bind_param('i,s,s,s', $date, $header, $summary, $content); $stmt->execute(); $stmt->close(); } else { echo "ERROR: SQL statement failure!"; echo "<a href='addnews.php'> -> OK</a>"; } $mysqli->close(); It looks fine to me, just can't see whats wrong lol! Within PHP I am attempting to insert some data into a MySQL table, however the value that needs to be stored inside the database field contains a semi-colon ; $q_options_data = "INSERT INTO mytable SET myfield = 'a:5:{s:13:\"administrator\";a:2:{s:4:\"name\";s:13:\"Administrator\";'"; I tried just escaping the ; with a \; but that didn't work I am using PHP 5.2.9 and MySQL 5.0.91-community Thanks, Chad I'm trying to create a INSERT query statement that makes use of the content of 'include' files. When I call up the include it simply echos it to the screen but I can't seem to capture the text string and actually make it function as part of the query. I've tried setting the include as a variable but again it only outputs to the screen. Help please! Here's a portion of the code... if (($_POST['sched_templates']) == "sched_TestingOnly.htm") { $query = include 'sched_TestingOnlyQuery1.htm'; } if (@mysql_query ($query)) { mysql_close(); // End adding Registration to the database header ('Location: redirectfile.htm'); exit(); } else { print "<p><b>Sorry, your entry could not be entered. } I have a form that users input user name, password, and email... all writes to csv fine. My problem is I need to concatenate a a string with a variable and some html code will preserving the html when written to the csv.. I need the csv to be this username,password,email,user,category,text with http://sub.domain..com/splash/,something,something here's what i have Code: [Select] <?php if($_POST['formSubmit'] == "Submit") $varUserName = $_POST['username']; $varPW = $_POST['PW']; $varEmail = $_POST['email']; { $fs = fopen("testcsv.csv","a"); fputcsv($fs, array($varUserName,$varPW,$varEmail,"user","title",",category","some text '<a href="http://$varUserName.url.com/splash/>site.com</a>',)); fclose($fs); exit; } ?> and of course I'm getting Parse error: syntax error, unexpected T_STRING, expecting ')' on line 9 I am trying to modify a header redirect similar to this: header("Location: index.html"); exit; I would like to do something like this: header("Location: index.html?user=' . $user . '"); exit; Ofcourse, this does not work. Any hints? Hello. I have a simple enough code that takes information from one table and drops it into another. This is great, but I have 2a new complexities that I have been unable to code correctly. A. 'lastname', 'firstname' on table1 need to be combined into 'name' How & Where do I combine these strings and then pass them? Code: [Select] <?php include('dbconfig.php'); // Make a MySQL Connection mysql_connect("localhost", "$user", "$password") or die(mysql_error()); mysql_select_db("$database") or die(mysql_error()); $result = mysql_query( "INSERT INTO table2 (lastname, firstname, email) SELECT lastname, firstname, email FROM table1 WHERE email='someemail@gmail.com' ") or die(mysql_error()); ?> Hello! How do I insert the content of two different variables into the SAME field in a table? Let's say, that variable A contains the word "blue" and variable B contains the word "sky". How do I "merge" these to variables into only one variable containing the words "Blue Sky"? Best regards Morris |
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