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PHP - Php Sql Select Error In Code Please Help
please can you help .. I'm passing a value from another page using $_SESSION and trying to then use it in a SELECT statement in SQL think I'm missing something ... if hard code the value it works , and I have also checked at the other end that the variable is being assigned ...
code is (its the = $_SESSION['g_district'] "); that's causing the issue
I'm getting Parse error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting '-' or identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) in /Applications/XAMPP/xamppfiles/htdocs/mytimekeeper/add/add.php on line 170 in the browser
<div class="row s2 m2 l2">
<div>
<label for="course_id">Choose the course raced:</label>
<select name="course_id">
<?php
// query to create course_id dropdown
$resultset = $conn->query("SELECT `course_id`,`course_name` FROM `tbl_courses` WHERE `district`= $_SESSION['g_district'] ");
while($row = $resultset->fetch_assoc())
{
echo "<option value='{$row[course_id]}'>{$row['course_name']}</option>";
}
?>
</select>
Similar TutorialsHello: I am trying to select all the records from a table, and display them. Having an issue... This is the code: Code: [Select] <ul> <?php $query = mysql_query("SELECT id,mySectionTitle FROM myWebSiteData"); while($menuData = mysql_fetch_array($query)) { <li><a href="a_websiteData.php?id=" .echo $menuData['id'] ."">echo $menuData['mySectionTitle']</a></li> } ?> </ul> This is the error: Code: [Select] Parse error: syntax error, unexpected '<' in /html/admin/a_websiteData.php on line 91 line 91 is the one that starts with "<li>" What am I missing, or what would be the best way to do this? Hi: I have another small issue like the one I just posted about. I am trying to get the full state name based upon the state abbreviation. Like this: Code: [Select] <?php $query=mysql_query("SELECT full_state FROM zip_codes WHERE abbr_state = abbr_state") or die("Could not get data from db: ".mysql_error()); $full_state=$result['full_state']; ?> ... State Name: <input type="text" name="full_state" value="<?php echo $full_state; ?>" /><br /> [CODE] No errors, but also no "full_state" appears. (It is pulling in the correct "abbr_state") What am I missing here? I'm trying to figure out some pagination on a left joined query, and not having any luck. The query seems to be pulling the entire database, so I went into the query's WHERE and added the table name "descriptors." Now I'm getting this error with the below query: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /data/21/2/40/160/2040975/user/2235577/htdocs/leafsearch4.php on line 35 also: Unknown column 'descriptors.leaf_shape' in 'where clause' I thought this was syntax, but maybe I'm missing something. Code: [Select] $query = "SELECT COUNT(*) as num FROM descriptors LEFT JOIN plantae ON (descriptors.plant_id = plantae.plant_name) WHERE `descriptors.leaf_shape` LIKE '%$select1%' AND `descriptors.leaf_venation` LIKE '%$select3%' AND `descriptors.leaf_margin` LIKE '%$select4%'"; $total_pages = mysql_fetch_array(mysql_query($query)); PHPBB3 does it. I crafted it on my own forum too. Why not have it here? Code is being posted daily. https://www.phpbb.co...655675#p5655675 See it in action? That would be usefull here instead of dragging and highlighting our mouse over the code and pressing CTRL+C. If IPB has a custom BBCODE manager, you could just use the code in their topic @Phillip. Edited by Monkuar, 25 January 2015 - 12:19 PM.
Example Code: if FirstImage=1 is absent, continue with FirstImage=0 Thanks for your idea and effort. I have searched this forum as well as over 200 other forums and have not found the answer that is specific to my question. I have shortened my code drastically to assist in resolving this quickly -
I have a search form that has criteria for the search criteria with "virtual" "columns" in an array but it's not working. If I search one column at a time it works just fine but when I try to search 8 columns with one select I get the following error: SELECT Error: Unknown column 'achievements' in 'where clause'.
When a user selects search in Achievements, I need it to look at all 8 columns that are associated with achievements and bring back the results that match - the same as if the user selects search in Associations, I need it to look at all 5 columns and bring back the results that match.
My shortened code is as follows:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Search</title>
</head>
<body>
<form name="search" action="" method="POST">
<p>Search:</p>
<p>
Achievements/Associations: <input type="text" name="find1" /> in
<Select NAME="field1">
<Option VALUE="achievements">Achievements</option>
<Option VALUE="associations">Associations</option>
</Select>
<br><br>
Secondary Education: <input type="text" name="find2" /> in
<Select NAME="field2">
<Option VALUE="edu1sectype">Highest Certificate Attained</option>
<Option VALUE="edu1secname">Highest Grade Passed</option>
<Option VALUE="edu1secinst">Name of High School</option>
<Option VALUE="edu1secdate">Date Completed</option>
<Option VALUE="edu1secinsttyp">Type of Institution</option>
<Option VALUE="subjects">Subjects</option>
</Select>
<br><br>
<input type="hidden" name="searching" value="yes" />
<input type="submit" name="search" value="Search" />
</p>
</form>
<?php
$searching = $_POST['searching'];
$find1 = $_POST['find1'];
$field1 = $_POST['field1'];
$find2 = $_POST['find2'];
$field2 = $_POST['field2'];
if ($searching =="yes")
{
echo "<br><b>Searched For:</b> $find1 $find2<br>";
echo "<br><h2>Results</h2><p>";
//If they did not enter a search term we give them an error
// Otherwise we connect to our Database
include_once "connect_to_mysql.php";
mysql_select_db("table_name") or die(mysql_error());
// We preform a bit of filtering
$find = strtoupper($find);
$find = strip_tags($find);
$find = trim($find);
$find = mysql_real_escape_string($find);
$field = mysql_real_escape_string($field);
$data = mysql_query("SELECT * FROM table_name
WHERE upper(".$field1.") LIKE '%$find1%'
AND upper(".$field2.") LIKE '%$find2%'
") or die("SELECT Error: ".mysql_error());
$result = mysql_query("SELECT * FROM table_name
WHERE upper($field1) LIKE '%$find1%'
AND upper($field2) LIKE '%$find2%'
") or die("SELECT Error: ".mysql_error());
$num_rows = mysql_num_rows($result);
echo "There are $num_rows records:<br>";
echo '<center>';
echo "<table border='1' cellpadding='5' width='990'>";
// set table headers
echo "<tr><th>Reference</th>
<th>First Name</th>
<th>Last Name</th>
</tr>";
//get images and names in two arrays
$name= $row["name"];
$surname= $row["surname"];
$achieve1 = $row["achieve1"];
$achieve2 = $row["achieve2"];
$achieve3 = $row["achieve3"];
$achieve4 = $row["achieve4"];
$achieve5 = $row["achieve5"];
$achieve6 = $row["achieve6"];
$achieve7 = $row["achieve7"];
$achieve8 = $row["achieve8"];
$assoc1 = $row["assoc1"];
$assoc2 = $row["assoc2"];
$assoc3 = $row["assoc3"];
$assoc4 = $row["assoc4"];
$assoc5 = $row["assoc5"];
$edu1sectype = $row["edu1sectype"];
$edu1secinst = $row["edu1secinst"];
$edu1secname = $row["edu1secname"];
$edu1secdate = $row["edu1secdate"];
$edu1secinsttyp = $row["edu1secinsttyp"];
$subject1 = $row["subject1"];
$subject2 = $row["subject2"];
$subject3 = $row["subject3"];
$subject4 = $row["subject4"];
$subject5 = $row["subject5"];
$subject6 = $row["subject6"];
$subject7 = $row["subject7"];
$subject8 = $row["subject8"];
$compsoft1name = $row["compsoft1name"];
$compsoft2name = $row["compsoft2name"];
$compsoft3name = $row["compsoft3name"];
$compsoft4name = $row["compsoft4name"];
$compsoft5name = $row["compsoft5name"];
$compsoft6name = $row["compsoft6name"];
$achievements = array('achieve1', 'achieve2', 'achieve3', 'achieve4', 'achieve5', 'achieve6', 'achieve7', 'achieve8');
$associations = array('assoc1', 'assoc2', 'assoc3', 'assoc4', 'assoc5');
$subjects = array('subject1', 'subject2', 'subject3', 'subject4', 'subject5', 'subject6', 'subject7', 'subject8' );
$compsoft = array('compsoft1name', 'compsoft2name', 'compsoft3name', 'compsoft4name', 'compsoft5name', 'compsoft6name');
while ($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td ALIGN=LEFT>" . $row['id'] . "</td>";
echo "<td ALIGN=LEFT>" . $row['name'] . "</td>";
echo "<td ALIGN=LEFT>" . $row['surname'] . "</td>";
echo "</tr>";
}
echo "</table>";
//This counts the number or results - and if there wasn't any it gives them a little message explaining that
$anymatches=mysql_num_rows($data);
if ($anymatches == 0) {
echo "Sorry, but we can not find an entry to match your query";
}
}
?>
</body>
</html>
Any assistance will be greatly appreciated as I have been working on this website for the past 4 months which has totalled over 150 pages and this is one of the last pages left to program and it's taken 6 days to get to this search page to this point.
Hi all, Currently my dob_day is not showing any value, even though there is a value in the database, it seems like it is unable to retrieve the value. No value was shown in the <select> box. Do you guys have any idea? Thanks Code: [Select] <?php $query = "SELECT name, nric, gender, picture, dob_day FROM tutor_profile WHERE tutor_id = '" . $_GET['tutor_id'] . "'"; $data = mysqli_query($dbc, $query) or die(mysqli_error($dbc)); // The user row was found so display the user data if (mysqli_num_rows($data) == 1) { $row = mysqli_fetch_array($data); if ($row != NULL) { $name = $row['name']; $nric = $row['nric']; $gender = $row['gender']; $old_picture = $row['picture']; $dob_day = $row['dob_day']; } else { echo '<p class="error">There was a problem accessing your profile.</p>'; } //HTML CODING <label for="dob" class="label">Date Of Birth</label> <select id="dob_day" name="dob_day" value="<?php if (!empty($dob_day)) echo $dob_day; ?>"> <option> <option>01</option> <option>02</option> <option>03</option> <option>04</option> <option>05</option> <option>06</option> <option>07</option> <option>08</option> <option>09</option> <option>10</option> <option>11</option> <option>12</option> <option>13</option> <option>14</option> <option>15</option> <option>16</option> <option>17</option> <option>18</option> <option>19</option> <option>20</option> <option>21</option> <option>22</option> <option>23</option> <option>24</option> <option>25</option> <option>26</option> <option>27</option> <option>28</option> <option>29</option> <option>30</option> <option>31</option> </select> ?> Hi,
I'm really at a loss here.
I have queried a table to get an option list, which returns what I expect but I need to then add it to some current code instead of the fixed option list presented.
The syntax is beyond me.
Please see attached.
Any pointers would be great.
Attached Files
OptionList.txt 1.61KB
7 downloads Howdy everyone, please i need help changing a php coded form from a checkbox to a select menu. Here's the form. <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" > <table class="dtable2"> <tr><th colspan="5">Enter a domain name:</th></tr> <tr><td colspan="5"><center>www.<input name="domain" type="text" size="35" /></center></td></tr> <tr><th colspan="5">Select an extension:</th></tr> <tr> <?php $i = 0; foreach ($this->serverList as $value) { if ($value['check'] == true) $checked=" checked "; else $checked = " "; echo '<td><input type="checkbox" name="top_'.$value['top'].'"'.$checked.'/>.'.$value['top'].'</td>'; $i++; if ($i > 4) { $i = 0; echo '</tr><tr>'; } } ?> </tr> </table> <center><input type="submit" name="submitBtn" class="sbtn" value="Check" /></center> </form> <?php I'll really appreciate your help. I have a search function in php where the text characters are matched to characters in a tables field--- works perfectly.... I need to make the input box have a droplist of words from database, this is also easy for me to do. the problem here is there is no definitive value! the options list always outputs a blank in the url--- its supposed to search a matching value and then output the matching value to url... Here is the droplist code: $output['RATESTITLE']='<input class="inputbox" type="text" size="24px" name="ratestitle" value="'.$sch->filter['ratestitle'].'" onfocus="if (this.value ==\''.$output['LANGUAGE_SEARCH_RATESTITLE'].'\') {this.value = \'\'}" />'; this outputs a input text box--- i want to have a droplist of options to populate this text box... If you must know this is the third day im at it... Hello, first time poster.. I've looked the web over for a long time and can't figure this one out. - Below is basic code that successfully checks MySQL for a match and displays result. I was debugging and forced the "height" and "width" to be 24 and 36 to make sure that wasn't the problem. That's good.. - I'd like to give the user ability to select width and height from a form.. and have it do an onchange this.form.submit so the form can be changing as fields are altered (thus the onchange interaction) - In a normal coding environment I've done this numerous times with no "Page cannot be displayed" problems. It would simply change one select-option value at a time til they get down the form and click submit... but in WordPress I'm having trouble making even ONE single onchange work! - I've implemented the plugins they offer which allows you to "copy+paste" your php code directly into their wysiwyg editor. That works with basic tests like my first bullet point above. - I've copied and pasted the wordpress url (including the little ?page_id=123) into the form "action" url... that didn't work... tried forcing it into an <option value=""> tag.. didn't work. I'm just not sure. I've obviously put xx's in place of private info.. Why does this form give me Page Cannot Be Displayed in WordPress every time? It won't do anything no matter how simple.. using onchange.. Code.. $con = mysql_connect("xxxx.xxxxxxx.com","xxxxxx","xxxxx"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("xxxxxx", $con); $myprodwidth=24; $myprodheight=36; $result = mysql_query("SELECT * FROM product_sizes WHERE prodwidth='$myprodwidth' and prodheight='$myprodheight'"); while($row = mysql_fetch_array($result)) { echo $row['prodprice']; } mysql_close($con); <form method="post" action=""> <select name="myheight" onchange="this.form.submit();"> <option selected="selected" value="">select height</option> <option value="xxxxxxxxx.com/wordpress/?page_id=199&height=36">36</option> <option value="xxxxxxxxx.com/wordpress/?page_id=199&height=36">48</option> </select> Fairly new to PHP and have the following code which should simply output into 3 columns - I have checked the query which does work - so I dont understand why I get this error mysql_fetch_array() expects parameter 1 to be resource,
// Create connection
// Check connection
$sql = “SELECT * FROM duration_matched WHERE groupm = '”.$coincidence ."’";
// set up loop counter
// start table and first tr ’;
while ($row=mysql_fetch_array($result)) {
’;
// always output the td
’;
// then close off the last row and the table ’ . $row[‘image’] . ‘ ’; $conn->close(); Why doesn't this code work? I can't figure it out. Database information: DB: abelltx table: homework Fields in table: id, pid, uid, name, url, approved, comment, SubmissionTime <?PHP $link = mysql_connect('xxxxxx', 'xxxxx', 'xxxxxxx') or die(mysql_error()); mysql_select_db("abelltx", $link) or die(mysql_error()); $uid=abelltx $query = "SELECT DISTINCT name FROM homework ORDER BY id ASC"; $result = query_db($query); while ($myrow = mysql_fetch_array($result)) { echo $myrow[name].", "; } ?> I get the following error: "Parse error: syntax error, unexpected T_VARIABLE in D:\Hosting\5366560\html\test.php on line 6" I have a page that displayes a table from my database. What I'm attempting to do is create a page where you can delete rows of that table. I'm creating 3 pages 1( select the student or row to delete) 2( comfirm whether you want to delete that student) and 3( where the student has been deleted) I'm already having trouble with my first page. This is my code to select the row or student you want to delete: <?php Code: [Select] <?php include 'includes/config.php'; $delete_sql ="SELECT sno,cname, sname FROM student ORDER BY sno DESC"; $delete_query = mysql_query($delete_sql) or die(mysql_error()); $rsDelete = mysql_fetch_assoc($delete_query); ?> <html> <head> </head> <body> <p> select student to delete </p> <?php do {?> <p><a href="deletecomfirm.php?sno= <?php echo $rsDelete['sno']; ?>"> <?php echo $rsDelete['cname']; ?> by <?php echo $rsDelete['fname']; ?> </a> </p> <?php} while ($rsDelete = mysql_fetch_assoc($delete_query)) ?> </body> </html>?> I keep getting this error: Parse error: syntax error, unexpected $end in C:\Program Files\EasyPHP-5.3.3\www\Project\selectdelstudent.php on line 32 I'm assuming this error indicates an improper closing tag or something? Though I can't see my mistake. Also if you have any tips on improving my code feel free to comment so I have been making a php chat service, and when i use this code:
Welcome <?php echo $_COOKIE["username"]; ?><br>
chatroom: <?php echo $_COOKIE["chatroom"]; ?>
<?php
$phpfile=$_COOKIE["chatroom"];
$chpage = $phpfile .".php";
$chtext = $phpfile .".txt";
$filename = $_COOKIE["chatroom"] . ".txt";
$myfile = fopen($filename, "r") or die("Unable to open file!");
fopen($myfile);
fclose($myfile);
include $chtxt;
?>
<form action="post.php" method="post">
message: <input type="text" name="msg"><br>
<input type="submit">
</form>
I get a 403 error. I does anyone know that is wrong with the code?
there are more files that set the cookies but they work fine. I can't find anything wrong. please help.
$bbreplace = array ( '/(\[[Bb]\])(.+)(\[\/[Bb]\])/', '/(\[[Ii]\])(.+)(\[\/[Ii]\])/', '/(\[[Uu]\])(.+)(\[\/[Uu]\])/', '/(\[[Ss]\])(.+)(\[\/[Ss]\])/', '/(\[url=http://)(.+)(\])(.+)(\[\/url\])/', '/(\[img\])(.+)(\[\/img\])/', '/(\[IMG\])(.+)(\[\/IMG\])/', '/(\[COLOR=)(.+)(\])(.+)(\[\/COLOR\])/', '/(\[color=)(.+)(\])(.+)(\[\/color\])/', '/(\[SIZE=)(.+)(\])(.+)(\[\/SIZE\])/', '/(\[size=)(.+)(\])(.+)(\[\/size\])/', '/(\[list\](.+)(\[\/list\])/', '/(\[list=1\](.+)(\[\/list\])/', '/(\[lLIST\](.+)(\[\/LIST\])/', '/(\[LIST=1\](.+)(\[\/LIST\])/', '/(\[*\](.+)/' ); $bbreplacements = array ( '<b>\\2</b>', '<i>\\2</i>', '<u>\\2</u>', '<s>\\2</s>', '<a href="\\2">\\4</a>', '<img src="\\2">', '<img src="\\2">', '<font color="\\2">\\4</font>', '<font color="\\2">\\4</font>', '<font size="\\2">\\4</font>', '<font size="\\2">\\4</font>', '<ul>\\2</ul>', '<ol>\\2</ol>', '<ul>\\2</ul>', '<ol>\\2</ol>', '<li>\\2</li>' ); $PostText = preg_replace($bbreplace, $bbreplacements, $PostText); This is what I have to do BB Codes so Far. Now when I go to view the thread I get: Warning: preg_replace() [function.preg-replace]: Compilation failed: missing ) at offset 25 in /home/rayth/public_html/forum/viewthread.php on line 95 Line 95: $PostText = preg_replace($bbreplace, $bbreplacements, $PostText); From testing I didn't get this error untill I added the tags I have a website for ordering business cards, and have encountered an error with either the HTML or PHP coding. The website has a form with several text boxes and drop-down menus to fill in the order form, and at the end after clicking the "submit" button, it generates an e-mail through a PHP script that sends the order form to the printer of the business cards. For some reason, whenever a user has a "job title" with the word "specialist" in it, the e-mail never gets sent. I removed all job titles from the one drop-down menu, to try and fix the problem, but now when users type "specialist" in the comments box at the bottom of the form, that blocks the e-mail from being sent as well. I have tried looking online for possible reasons why this one word would cause such a problem, but have not found the answer yet. Does anyone on this forum know why the word "specialist" would cause this issue, or know where I can look for a solution? If you can help me, that would be greatly appreciated. Thanks! I'm getting an error with this code:
<?php I'm in the mist of doing a login script for a new site. I have the script working, just as long as you get it right the first time, because after that, it seems to hang on the delay. Script is below
$("#login_button").click(function() {
$.post("inc/login.php", $("#loginForm").serialize(), function(data) {
var login = data.login;
var message = data.message;
if(login == true) {
$("#login").empty().delay(5000).fadeOut('slow');
}
if(login == false) {
$("#login_window").css('visibility','hidden');
$("#message").css('visibility','visible').html(message);
$("#message").delay(4000).queue(function() {
$("#message").empty().css('visibility','hidden');
$("#login_window").css('visibility','visible');
});
}
}, "json");
});
Basically, its supposed to show the returned error message for 4 seconds, then show the login form again. I'm using a regular button, not the submit type button. I don't have any errors in firebug. Anyone have any ideas?
(i know about show and hide, but those caused entirely to many problems).
Edited by richei, 14 May 2014 - 11:54 PM. Hey all, Basically I have this code which technically should work, but I have put an error in the logic somewhere and am really struggling to find it, I've been going over it for about an hour now :/ The error is that in the second part (// Enter them into the activities database if they're not already there), it won't enter them, and it won't show the word 'Randomevent1' either, so clearly somehow I have put in something to prevent it performing that step. Can anybody help me out? I'd be amazingly grateful, I have no clue how I've screwed it up. if(isset($_POST['submit'])){ $ownerid = $_SESSION['id']; // If completed = Y give an error $completeyn = "SELECT completed FROM activities WHERE playerno='$ownerid' AND activityno = '1'"; $completecheck=mysql_query($completeyn) or die(mysql_error()); while($row = mysql_fetch_array( $completecheck )) { if($row['completed'] == 'Y'){ echo 'Oops, you\'ve already done this twice today!';} else { // Enter them into the activities database if they're not already there $stepno2 = "SELECT playerno, timesdone FROM activities WHERE playerno='$ownerid' AND activityno = '1'"; $stepnoanswer2=mysql_query($stepno2) or die(mysql_error()); $num_rows2 = mysql_num_rows($stepnoanswer2); echo $num_rows2; if($num_rows2 == '0'){ $putintodb2 = mysql_query("INSERT INTO activities (playerno, activityno, timesdone) VALUES ('$ownerid', '1', '1')") or die("Error: ".mysql_error()); echo 'Randomevent1'; }else{ // If they are already there update their stepcount $updatestepcount2=("UPDATE activities SET timesdone=timesdone+'1' WHERE playerno='$ownerid' AND activityno = '1'"); $newstepcount2=mysql_query($updatestepcount2); echo 'Randomevent2';} // If this new stepcount is equal to 2, set completed to Y $checkstep = "SELECT timesdone FROM activities WHERE playerno='$ownerid' AND activityno = '1'"; $checkstepresult=mysql_query($checkstep) or die(mysql_error()); while($row = mysql_fetch_array( $checkstepresult )) { if($row['timesdone'] == '2'){ echo $row['timesdone']; $updatestepcount22=("UPDATE activities SET completed = 'Y' WHERE playerno='$ownerid' AND activityno = '1'"); $newstepcount22=mysql_query($updatestepcount22); } }}}} |
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