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PHP - Sql Insert Statement Is Not Working In Php Page.
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> Similar TutorialsI 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 $matches[1] How can I do that? Hi, I would like to know how to write Insert data into a table statement where some of the data is coming from another table? I tried using Insert select statement but did not work. Here is what I am trying.... 1.........$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'm having issues with the following: Code: [Select] <?php session_start(); $_SESSION['username']=$_POST['username']; $_SESSION['password']=$_POST['password']; if($_SESSION['username']=="username" && $_SESSION['password']=="password"){ if($_GET['product']=="add"){ $content.=' <p><label>Product Name:</label> <input type="text" name="product_name" size="30" /> <label>Product Price:</label> <input type="text" name="product_price" size="5" /> </p> <p><label>Product Category:</label> <input type="text" name="product_category" size="30" /></p> <p><label>Product Link:</label> <input type="text" name="product_link" size="30" /></p> <p><label>Product Image:</label> <input type="text" name="product_image" size="30" /></p> <p><label>Product Tag:</label> <input type="text" name="product_tag" size="30" /></p> <p><label>Product Keywords:</label> <input type="text" name="keyword" size="30" /></p> <p><label>Product Features:</label><br /> <textarea name="product_features" rows="10" cols="60"></textarea> </p> <p><label>Product Pros:</label><br /> <textarea name="product_pros" rows="5" cols="30"></textarea> </p> <p><label>Product Cons:</label><br /> <textarea name="product_cons" rows="5" cols="30"></textarea> </p> <p><label>Product Description:</label><br /> <textarea name="product_description" rows="10" cols="60"></textarea> </p> <p><label>Product Notes:</label><br /> <textarea name="product_notes" rows="5" cols="30"></textarea> </p> '; $logout='<div><a href="./acp_admincp.php?log-out">Log-Out</a></div>'; } elseif($_GET['product']=="view"){ } else{ $content.=' <a href="./admincp.php?product=add">Add New Product</a> <br /> <a href="./admincp.php?product=view">View Products</a> '; } } elseif(isset($_GET['log-out'])){ session_start(); session_unset(); session_destroy(); header("Location: ./admincp.php"); } else{ $content=' <form action="./admincp.php" method="post"> <p><label>Username:</label> <input type="text" name="username" size="30" />'; $content.='</p> <p><label>Password:</label> <input type="password" name="password" /></p>'; $content.='<p><input type="submit" value="Submit" name="Submit" /></p> </form>'; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en"> <head> <base href="http://ghosthuntersportal.com/" /> <title>Ghost Hunter's Portal - Admin Control Panel</title> <meta http-equiv="content-type" content="text/html; charset=utf-8" /> <meta name="verify-v1" content="" /> <meta name="keywords" content="ghost, hunters, hunter, ghosts, spirit, spirits, paranormal, investigation, investigator, investigators, k2, emf, meter, kii" /> <meta name="description" content="Ghost Hunters Potal. Parnormal research equipment store." /> <meta name="author" content="Andrew McCarrick" /> <meta name="robots" content="index, follow" <link rel="stylesheet" type="text/css" href="style.css" /> </head> <body> <img src="./logo.png" alt="Ghost Hunter's Portal Admin Control Panel" /> <br /> <div style="color: #AA44AA; font-size: 26px; margin-top: -30px; margin-left: 125px;">Admin Control Panel</div> <?php echo $logout; echo $content; ?> </body> </html> I can log-in, and get to the page with the two links on it. However, once I click one of the links it falls back to the log-in page, and it ends up being a never ending loop. It's doing this: Log-In --> Page with links ---> Log-In page again Should be doing this: Log-In --> Page with links --> Add Product page or View Products page I can never get into the the actual sub page. Just to be clear, the address bar actually shows product=add or product=view, but it still shows the log-in page. 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');"; 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.
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 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!"; ?> 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 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=315503.0 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. } Hello guys, I am having an issue with my if statement. Basically what I am trying to do is when there are no photos uploaded then enter default image. Like I said doesn't look like my if stmt is working at all.. Any ideas? foreach (glob("temp_photo/$subn*.*") as $filename) { //echo "$filename<br>"; $string = $filename; $filename2 = basename($string); if($filename2 == ""){ $filename2="/images/default.jpg"; } // Insert ad details into mysql $query = "INSERT INTO ads (ad_type,ad_title,ad_body,ad_category,ad_photo,ad_member,ad_status,ad_city,ad_state,ad_zip)VALUES('".$ad_type."','".$ad_title."','".$ad_body."','".$catid."','".$filename2."','".$vname."','".$ad_status."','".$city."','".$state."','".$zip."')"; // Error checking to see if the query was successful $result = mysql_query($query) or die (mysql_error()."in <br>$query" ); $query_rsGetOrderDetails = "SELECT * FROM hd_order WHERE order_by = '$usernameLoggedin' ORDER BY order_id DESC LIMIT 3"; Any ideas? Absolutely no idea why my IF statement's playing up - needs some fresh eyes I think! if ($postcheck != $post){ if ($x == "0"){ echo $title; echo "<br>"; $x = "1"; $y = $y + 1; } echo "Change Post to: "; echo $postcheck; echo " from "; echo $post; echo "<br>"; $result2 = mysql_query("UPDATE items SET Post='$postcheck' WHERE ID='$id'") or die(mysql_error()); } When it returns $postcheck and $post, in this case, they're coming out as the same thing (on one record - no others!) - $post is retrieved from a database, and $postcheck is calculated. Both are numbers. Any ideas?! Hello Everyone, From time to time, I get into a situation where my if statements do not give the desired results. I always have to play around with them to get the things working. I'm in same situation right now. I'm guessing that I do not completely understand the logic. If someone can take a look at the code below and tell me what's wrong, it will be greatly appreciated. if($_SESSION['account_type'] != 'internal admin'){ die('<font color="red"><b>Error:</b> This page can only be accessed by internal or super admin.</font>'); break; } elseif($_SESSION['account_type'] != 'super admin'){ die('<font color="red"><b>Error:</b> This page can only be accessed by internal or super admin.</font>'); break; } I have echoed out the $_SESSION['account_type'] and I'm getting "internal admin" but for some reason, I'm getting the error "Error: This page can only be accessed by internal or super admin." Thanks in advance for your help. Jatt Surma Ok, a bit of a background into my code, It runs off of 2 different passing variables, I have it set up with multiple if() statements to do certain things based on the inputs... Here is a snipet of my code if ($_REQUEST['postal'] != "" && $_REQUEST['lbs'] != "") { $postal = $_REQUEST['postal']; $lbs = $_REQUEST['lbs']; $type = $_REQUEST['type'] if ($type == "splist") { $ztype = "cansp_zone" ; } else { $ztype = "canmp_zone"; } Obviously not the whole thing, Now what you see is line 18-26 of a 216 line document. The nested if statement is what is causing the problem, I get a "syntax error, unexpected T_IF" on line 22, the "if ($type)" line. Now theoretically, if I did not pass the variables (in which case the primary if statement would be false) It should bypass this whole section, however its not. Any ideas? I am thinking its not actually doing the verification and outputting the error because type is not defined. i have been trying to figure this out for the last hour with no luck i am hopeing some one here can help me the problem is well i really dont know what the problem is but the code should check is the passwords match if they dont it will tell you to reenter your password if(isset($_POST['Submit'])) //isset open { $pass1=$_POST['pass1']; $pass2=$_POST['pass2']; if($pass1 != o) //pass open { echo"<style type='text/css'> <!-- body { background-color: #FFFFFF; } body,td,th { color: #000000; } --> </style></head> <body> <table width='100%' border='1'> <tr> <td><center>Your Password Did Not Match Please Reenter Your Passwrods</center></td> </tr> <tr> <td><form id='form1' name='form1' method='post' action=''> <label>Password <input type='text' name='pass1' id='pass1' /> </label> </td> </tr> <tr> <td> <label>Enter Password Again <input type='text' name='pass2' id='pass2' /> </label> </td> </tr> <tr> <td> <label> <center><input type='submit' name='Submit' id='Submit' value='Submit' /></center> </label> </form> </td> </tr> </table> </body>"; die(); //pass colse } else //pass else open { echo" <form name='redirect'> <center> <font face='Arial'><b>Your Account Has Been Validated You Will Now Be Redirect To The Admin Home Page in<br><br> <form> <input type='text' size='3' name='redirect2'> </form> seconds</b></font> </center> <script> <!-- //change below target URL to your own var targetURL='/test1' //change the second to start counting down from var countdownfrom=5 var currentsecond=document.redirect.redirect2.value=countdownfrom+1 function countredirect(){ if (currentsecond!=1){ currentsecond-=1 document.redirect.redirect2.value=currentsecond } else{ window.location=targetURL return } setTimeout('countredirect()',1000) } countredirect() //--> </script>"; //pass else close } //isset close } what i'm trying to do, is have it so a staff member can mange users i came up with these, but it dosn't show on the page :S if ( isadmin ( $_SESSION['user_id'] ) ): ?> <br /> It seems that you're an admin. You may <a href="manage_users.php" title="manage users">manage users</a> or <a href="admin_settings.php" title="edit site settings">edit site settings</a>. <? if ( ismod ( $_SESSION['user_id'] ) ): ?> <br /> It seems that you're an mod. You may <a href="manage_users.php" title="manage users">manage users</a>. <?php endif; ?> Hi there im having trouble with my If statement in that its not doing what it is suppose to do! I have a do loop that and If statament that checks that resultp < results. If it is less then an echo is suppose to happen, but its not. Im getting notice errors: Undefined index: Class1- 4 but nothing is outputted. I checked the records in the database so the condition will happen but nothings happening. Help Code: [Select] <?php do { ?> <?php echo $row_resultp['Class1_totla']; echo ' Class 1'; echo'<br>'; echo $row_resultp['Class2_totla']; echo ' Class 2'; echo'<br>'; echo $row_resultp['Class3_totla']; echo ' Class 3'; echo'<br>'; echo $row_resultp['Class4_totla']; echo ' Class 4'; echo'<br>';echo'<br>';echo'<br>'; echo $row_results['Class1_totlas']; echo ' Class 1'; echo'<br>'; echo $row_results['Class2_totlas']; echo ' Class 2'; echo'<br>'; echo $row_results['Class3_totlas']; echo ' Class 3'; echo'<br>'; echo $row_results['Class4_totlas']; echo ' Class 4'; if ($row_resultp['Class1'] < $row_results['Class1']){ echo 'you need'; echo $row_resultp['Class1'] - $row_results['Class1']; echo 'more'; } if ($row_resultp['Class2'] < $row_results['Class2']){ echo 'you need'; echo $row_resultp['Class2'] - $row_results['Class2']; echo 'more';} if ($row_resultp['Class3'] < $row_results['Class3']){ echo 'you need'; echo $row_resultp['Class3'] - $row_results['Class3']; echo 'more'; } if ($row_resultp['Class4'] < $row_results['Class4']){ echo 'you need'; echo $row_resultp['Class4'] - $row_results['Class4']; echo 'more'; } ?> <?php } while ($row_resultp = mysql_fetch_assoc($resultp) && ($row_results = mysql_fetch_assoc($results))); ?> |
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