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Posts posted by ann

    version/install issues

    in PHP Installation and Configuration

    Posted

    Thanks for replying

     

    I hadn't realised #php -v and #php /var/www/html/php_info.php were also giving different versions of php!

     

    php5-cli was installed and removed with apt-get and apt-get claims it's not installed now...

    #apt-get remove php5-cli
Reading package lists... Done
Building dependency tree       
Reading state information... Done
Package php5-cli is not installed, so not removed
0 upgraded, 0 newly installed, 0 to remove and 4 not upgraded.

     

    I've tried again to delete all the php files, restarted apache and when I load php_info.php in a browser PHP Version 5.2.4-2ubuntu5.10 is still running?!?  but I can't use php from the command line because there's no executable.  (I have check that edits to php_info.php are shown in the browser).

     

    I need to be able to get rid of PHP Version 5.2.4 or reconfigure it --with-mysql.

    Any suggestions?

     

    Thanks

     

     

    version/install issues

    in PHP Installation and Configuration

    Posted

    Dear All

     

    Please can someone help?  I get different version of php used from the command line than I get with a browser.  I've tried removing all php with 'apt-get remove' and deleted everything php that I dared like...

      350  rm -rf /var/lib/php*

      351  rm -rf /usr/local/bin/php*

      352  rm -rf /usr/bin/php*

      353  rm -rf /etc/php*

      364  rm -rf /var/lib/dpkg/info/php*

      369  rm -rf /usr/share/phpmyadmin

      372  rm -rf /usr/share/php5

      375  rm -rf /usr/share/doc/phpmyadmin

      378  rm -rf /usr/share/doc/php5*

      381  rm -rf /usr/local/lib/php

      384  rm -rf /usr/local/include/php

      387  rm -rf /usr/lib/php5

      388  rm -rf /usr/bin/php

      389  rm -rf /etc/cron.d/php5

      429  rm -rf /usr/local/lib/php

     

    Then configured and installed php-5.3.3. From the command line I get php-5.3.3 with mysql...

     

    root# php /var/www/html/php_info.php 
<html>
<head>
</head>
<body>
<center>php info file</center><br><br>
phpinfo()
PHP Version => 5.3.3

System => Linux bicr-bioinf4 2.6.24-28-generic #1 SMP Wed May 26 23:34:09 UTC 2010 x86_64
Build Date => Aug 19 2010 12:05:17
Configure Command =>  './configure'  '--with-mysql'
<snip>

     

    but loading the same file in a browser I get 5.2.4 with no mysql...

     

    php info file


PHP Logo
PHP Version 5.2.4-2ubuntu5.10

System 	Linux bicr-bioinf4 2.6.24-28-generic #1 SMP Wed May 26 23:34:09 UTC 2010 x86_64
Build Date 	Jan 6 2010 21:42:59
Server API 	Apache 2.0 Handler
Virtual Directory Support 	disabled
Configuration File (php.ini) Path 	/etc/php5/apache2
Loaded Configuration File 	(none)
Scan this dir for additional .ini files 	/etc/php5/apache2/conf.d
PHP API 	20041225
PHP Extension 	20060613
Zend Extension 	220060519
Debug Build 	no
Thread Safety 	disabled
Zend Memory Manager 	enabled
IPv6 Support 	enabled
Registered PHP Streams 	zip, php, file, data, http, ftp, compress.bzip2, compress.zlib, https, ftps
Registered Stream Socket Transports 	tcp, udp, unix, udg, ssl, sslv3, sslv2, tls
Registered Stream Filters 	string.rot13, string.toupper, string.tolower, string.strip_tags, convert.*, consumed, convert.iconv.*, bzip2.*, zlib.* 
<snip>

     

    >:(

     

    I've stopped and started apache, closed and opened various browsers, so I know I'm looking at the right server and it's not chached.

     

    What am I missing?

     

    Many thanks

     

     

     

    root# uname -a

    Linux bicr-bioinf4 2.6.24-28-generic #1 SMP Wed May 26 23:34:09 UTC 2010 x86_64 GNU/Linux

    root# apache2 -v

    Server version: Apache/2.2.8 (Ubuntu)

    Server built:  Jun 18 2010 14:04:18

    root# php -v

    PHP 5.3.3 (cli) (built: Aug 19 2010 09:25:31)

    Copyright © 1997-2010 The PHP Group

    Zend Engine v2.3.0, Copyright © 1998-2010 Zend Technologies

     

     

    php array elements and javascript?

    in Javascript Help

    Posted

    Hi

     

    I've got a page with a number of dependent dropdown menus.  i.e. using car manufacturer/models as an example...If the user selects 'Ford' as a manufacturer javascript reloads the page with ?manufacturer=Ford in the url and php restricts the model dropdown to models made by Ford.  It works the other way too, if they select a model first the list of manufacturers is restricted.

     

    Things just got more complicated and  users may select up to 3 models from the same manufacturer.  If I turn the form element 'model' into an array how do I get javascript to work with it?

     

    I just need what to to with the call to the reload() function and the javascript. I can deal with php arrays.

     

    Many thanks

     

     

    Bits of my code for a single model...

     

    html

    <select name="make" onchange="reload(this.form,'select.php')" >
    <option value="0" selected></option>
    <option value="1">Ford</option>
    <option value="2">VW</option>
    <option value="4">Opal</option>
</select>
<select name="model" onchange="reload(this.form,'select.php')" >
    <option value="0" selected></option>
    <option value="4">fiesta</option>
    <option value="23">fabia</option>
    <option value="24">td9</option>
</select>

     

    php

    function ddmodel(){
#model
if (isset($_REQUEST['make']) and $_REQUEST['make']>0) {
	$mainquery="SELECT DISTINCT id,model FROM model where make=".$_REQUEST['make']
}
else {$mainquery="SELECT id,model from model"; }

$string= "<td><select name=\"model[".$n."]\" onchange=\"reload(this.form,'".$_SERVER['PHP_SELF']."')\" >";
$string.= "<option value=\"0\" selected></option>\n";
$ddsql_result = mysql_query($mainquery);
if(mysql_num_rows($ddsql_result)){
	while($ddrow = mysql_fetch_assoc($ddsql_result)){
		if ($_REQUEST['model']==$ddrow['id']) {
			$string.= "<option value=\"".$ddrow['id']."\" selected>".$ddrow['model']."</option>\n";
		}
	else {$string.= "<option value=\"".$ddrow['id']."\">".$ddrow['model']."</option>\n";}
}}
$string.= "</select></td>\n";
return $string;
}

function ddmake(){
#make
if (isset($_REQUEST['model']) and $_REQUEST['model']>0) {
	$mainquery="SELECT make.make,make.id from make left join model on (make.id=model.make) where model.id=".$_REQUEST['model'];
}
else {$mainquery="SELECT make,id from make order by make"; }

$string= "<td><select name=\"make\" onchange=\"reload(this.form,'".$_SERVER['PHP_SELF']."')\" >";
$string= "<option value=\"\"selected></option>\n";
$ddsql_result = mysql_query("$mainquery");
if(mysql_num_rows($ddsql_result)){while($ddrow = mysql_fetch_assoc($ddsql_result)){
	if ($_REQUEST['make']==$ddrow['id']) {
		$string.= "<option value=\"".$ddrow['id']."\" selected>".$ddrow['make']."</option>\n";
	}
	else {
		$string.= "<option value=\"".$ddrow['id']."\">".$ddrow['make']."</option>\n";
	}
}}
$string.= "</select></td>\n";
return $string;
}

     

    javascript

    function reload(form,selfid){
var str='?';
if (form.make && form.make.options[form.make.options.selectedIndex].value>0) {
	 str +='&make=' + form.make.options[form.make.options.selectedIndex].value;
}
if (form.model && form.model.options[form.model.options.selectedIndex].value>0) {
	str +='&model=' + form.model.options[form.model.options.selectedIndex].value;
}
self.location=selfid + str;
}

     

    [SOLVED] document.getElementById(" ") is null issue.

    in Javascript Help

    Posted

    Hi,

     

    I want the contents of a text box to change with the selection from a dropdown menu.  The script below works, but only if I include the line marked <!--why do I need this-->.  If I delete this line the page loads without error but I get document.getElementById("divSCO0") is null when I make a selection and no changing text.

     

    How can I get the script to work without the <!--why do I need this--> line?

    or

    Can I get divXXX not to display? I've got style="display:none" but it still shows!

     

    The number of select options and divs  will vary with database content.  Could the java script be written before I have the results of the database query? I guess I'd have to pass it both the total number of options and which option to display?

     

    Thanks for your time.

     

    <html>
<head>
<script type="text/javascript">
function checker(what){
	document.getElementById("divSCO0").style.display = "none";
	document.getElementById("divSCO1").style.display = "none";
	document.getElementById("divSCO2").style.display = "none";
if (what==1) {
	document.getElementById("divSCO1").style.display = "block";
}else if (what==2) {
	document.getElementById("divSCO2").style.display = "block";
}else {
	document.getElementById("divSCO0").style.display = "block";
}
} 
</script>
</head>

<body>

Select 
<select name="DDselect" onchange="checker(this.value)"><option value="0" selected></option>
<option value="1">1</option>
<option value="2">2</option>

<br><br>

<div id="divXXX" style="display:none"><textarea name="XXX" id="XXX" rows="0" cols="0"></textarea></div><!--why do I need this-->
<div id="divSCO0" style="display:block"><textarea name="SCO0" id="SCO0" rows="2" cols="30"></textarea></div>
<div id="divSCO1" style="display:none"><textarea name="SCO1" id="SCO1" rows="2" cols="30">Selection one from the dropdown</textarea></div>
<div id="divSCO2" style="display:none"><textarea name="SCO2" id="SCO2" rows="2" cols="30">Two selected</textarea></div>

</body>
</html>

    [SOLVED] query, order, group by issue.

    in MySQL Help

    Posted

    I don't have the language to make the question obvious.  I need the rows grouped on probe number and the groups ordered by the lowest pvalue in each group. 

     

    The lowest pvalue in the data set is

          c              3          0.0001

     

    so I want all the probe 3's first.  Having used the probe 3's, the lowest pvalue in the set is now...

          a              1          0.001

     

    so I want all the probe 1's next.  And if I'd used "WHERE pvalue<0.08" I'd get all the probe 2's next.

    (Within group ordering on pvalue would be nice but it's secondary to getting the groups in the right order)

     

    I found a page in the manual about Create Procedure or Function.  Maybe that's the way to go.  I'll go do some reading.

    [SOLVED] query, order, group by issue.

    in MySQL Help

    Posted

    Thanks, 

     

    I think that gives me the probe 1 group first, but I need probe 3 first because that group contains the lowest pvalue in the whole data set.

     

    If you imagine all these rows chucked in a pile I want to take from the pile the row with the lowest pvlaue,

    then take all the other rows with the same probe id,

    then from what's left take the lowest pvalue again, etc. 

     

    If it's possible?

     

     

    [SOLVED] query, order, group by issue.

    in MySQL Help

    Posted

    Hi

     

    Can some one tell me if this is possible in a single query and if so how would I go about it?

     

    If I've got a table like...

    experiment           probe         pvalue
       a               1           0.001
       a               2           0.7
       a               3           0.01
       b               1           0.7
       b               2           0.7
       b               3           0.0002
       c               1           0.7
       c               2           0.7
       c               3           0.0001

     

    and I want results like...

    experiment           probe         pvalue
       c               3           0.0001
       b               3           0.0002
       a               3           0.01
       a               1           0.001
       b               1           0.7
       c               1           0.7

     

    I can do this in two steps by...

        SELECT distinct(probe) FROM TABLE WHERE pvalue<0.05 order by pvalue;

    then using php to loop through the results ...

        SELECT * FROM TABLE WHERE probe='<each results from above query>' order by pvalue

    but it's very slow.

     

    I'll try and put it in words as 'I want all the rows for those probes which had a pvalue<0.05 in any experiment and I want the results grouped by probe and ordered by the lowest pvalue of each group of probes'.

     

    I can't even think of a search term that might help me solve this so any hints would be appreciated.

     

    Thanks

     

     

    MySQL Server version: 5.0.77

     

    Reload Function

    in Javascript Help

    Posted

    rhodesa

     

    I want to post variables from a form to 'test1.php' via your script.  I thought I'd set the variables as session cookies at the start of your script but there are no session cookies when test1.php runs.  It's probably the wrong approach anyway can you advise on the correct way to do this?

     

    Thanks

     

    My version of your script which reports that session cookies but test1.php doesn't

    <?
session_start();  // Start Session 
session_register('analysis'); 
$_SESSION['analysis'] = $_REQUEST['analysis'];
session_register('query'); 
$_SESSION['query'] = $_REQUEST['query'];
?>

<html>
   <head>
      <title>Testing Javascript</title>
      <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3/jquery.min.js"></script>
      <script type="text/javascript">
        $(function(){
          //This code will run once the DOM is done loading
          setInterval("$('#time').load('test1.php');",1000);
        });
      </script>
   </head>
   <body>
      <div id="time" style="border: 1px black solid; width: 750px;"></div>
   <? 
      echo $_SESSION['analysis']." analysis<br>";
      echo $_SESSION['query']." query<br>";
   ?>
   </body>
</html>

    [SOLVED] MYSQL_BOTH - column name identification

    in MySQL Help

    Posted

    Thanks Barand.  Don't think I asked the question very well and I probably should have asked under php. 

     

    What I was after is the array_keys function which gives an array of the column names from a query.  My code now looks like...

     

    $sql = mysql_query($query);
$columns= array_keys(mysql_fetch_array($sql, MYSQL_ASSOC)); 	

$sql = mysql_query($query);
while($row = mysql_fetch_array($sql, MYSQL_BOTH)){ 
     //work out which field matched and print that field
     for ($i=1; $i<sizeof($row); $i++) {
        if (preg_match('/'.$_REQUEST['search_key'].'/', $row[$i])){
           $row[$i]=preg_replace('/'.$_REQUEST['search_key'].'/', '<b>'.$_REQUEST['search_key'].'</b>', $row[$i]);  #make the search term bold
           echo $columns[$i]." ".$row[$i]."<br>";
        }
     }
}

     

    NOTE: if you're joining tables and those tables have columns with the same name (I had edit_date columns in two tables) the array_keys function will only give you the first instance of the name so the $columns array and the $row array end up different lengths.

    [SOLVED] mysql joins query.

    in MySQL Help

    Posted

    I may not fully understand your problem from the description you gave but maybe a sub query is what you're after?

     

    select cardtagid from cardtags where tagid='C';

     

    Should give you a result of 5,6,7,8,9,13. So a query for all tagid's that have a cardtag in the list 5,6,7,8,9,13 would be something like...

     

    Select distinct(tagid) from cardtags where cardtagid in (select cardtagid from cardtags where tagid='C');

     

    which should give you A, B, C, D. Then to get rid of the tagid you used as a parameter, in this case C, add and tagid!='C' to the query...

     

    Select distinct(tagid) from cardtags where cardtagid in (select cardtagid from cardtags where tagid='C') add and tagid!='C';

     

    Hope that helps.

     

    [SOLVED] MYSQL_BOTH - column name identification

    in MySQL Help

    Posted

    Hi

     

    I'm working on a 'quick', 'key word' based search for my database.  Say the key word is 'pool' and the sql search returns an array like...

    id : image_desc    : location        : photographer

    2  : mountain pool  : lake district  : Mr Swan

    8  : tower ballroom : blackpool    : Mrs Pen

    9  : beach ball        : skegness    : James Pool

     

    The code below will print...

    2 mountain pool

    8 blackpool

    9 James Pool

     

    What I'd like is...

    2 mountain pool (image_desc)

    8 blackpool (location)

    9 James Pool (photographer)

     

    I can do this by creating an array of all my column names and using $i to pull out the relative column id but is there a better way to do it?  I'm using MYSQL_BOTH so I already have the column name information I just don't know how to get at it.

     

    Thanks for your time.

    Ann

     

    		
while($row = mysql_fetch_array($sql, MYSQL_BOTH)){ 
     //work out which field matched and print that field
     for ($i=1; $i<sizeof($row); $i++) {
        if (preg_match('/'.$_REQUEST['search_key'].'/', $row[$i])){
           $row[$i]=preg_replace('/'.$_REQUEST['search_key'].'/', '<b>'.$_REQUEST['search_key'].'</b>', $row[$i]);  #make the search term bold
           echo $row[0]." ".$row[$i]."<br>";
        }
     }
}

    multiple versions and reconfigure

    in Linux

    Posted

    Hi

     

    I want to reconfigure php --with-mysql. I thought this would be as simple as take the configure command from the phpinfo() page and run it from the php installation dir changing the --without to --with for mysql. But, (isn't there always a but?) I'm running php-5.2.2 and I can only find install directories for php-4.3.7 and php-5.0.1. 

    Does anyone know why I appear to be using a different version to the ones installed?

    As I haven't got a /usr/software/apache/php-5.2.2/configure file how do I reconfigure for mysql?

     

    Cheers

    Ann

     

    OS: FC7 and a new install of about a month ago so I've no idea why I have two (or is it three) php versions.

    [root@localhost]# which php
/usr/bin/php

[root@localhost]# /usr/bin/php -v
PHP 5.2.2 (cli) (built: May  8 2007 08:15:05) 
Copyright (c) 1997-2007 The PHP Group
Zend Engine v2.2.0, Copyright (c) 1998-2007 Zend Technologies

[root@localhost]# locate php-5.2
[root@localhost]#

[root@localhost]# locate php | grep configure
/usr/software/apache/php-4.3.7/configure
/usr/software/apache/php-4.3.7/configure.in
/usr/software/apache/php-4.3.7/TSRM/configure.in
/usr/software/apache/php-4.3.7/Zend/configure.in
/usr/software/apache/php-4.3.7/ext/bcmath/libbcmath/configure
/usr/software/apache/php-4.3.7/ext/bcmath/libbcmath/configure.in
/usr/software/apache/php-4.3.7/ext/libpng/configure
/usr/software/apache/php-4.3.7/ext/mbstring/libmbfl/configure.in
/usr/software/apache/php-5.0.1/configure
/usr/software/apache/php-5.0.1/configure.in
/usr/software/apache/php-5.0.1/TSRM/configure.in
/usr/software/apache/php-5.0.1/Zend/configure.in
/usr/software/apache/php-5.0.1/ext/bcmath/libbcmath/configure
/usr/software/apache/php-5.0.1/ext/bcmath/libbcmath/configure.in
/usr/software/apache/php-5.0.1/ext/mbstring/libmbfl/configure.in
/usr/software/apache/php-5.0.1/win32/build/configure.tail
[root@localhost]#

     

    The start of my phpinfo()...

     

    PHP Version 5.2.2

System 	Linux localhost.localdomain 2.6.21-1.3194.fc7 #1 SMP Wed May 23 22:35:01 EDT 2007 i686
Build Date 	May 8 2007 08:15:45
Configure Command 	'./configure' '--build=i386-koji-linux-gnu' '--host=i386-koji-linux-gnu' '--target=i386-redhat-linux-gnu'
'--program-prefix=' '--prefix=/usr' '--exec-prefix=/usr' '--bindir=/usr/bin' '--sbindir=/usr/sbin' '--sysconfdir=/etc' 
'--datadir=/usr/share' '--includedir=/usr/include' '--libdir=/usr/lib' '--libexecdir=/usr/libexec' '--localstatedir=/var' 
'--sharedstatedir=/usr/com' '--mandir=/usr/share/man' '--infodir=/usr/share/info' '--cache-file=../config.cache' '--with-libdir=lib' 
'--with-config-file-path=/etc' '--with-config-file-scan-dir=/etc/php.d' '--disable-debug' '--with-pic' '--disable-rpath' '--without-pear' 
'--with-bz2' '--with-curl' '--with-exec-dir=/usr/bin' '--with-freetype-dir=/usr' '--with-png-dir=/usr' '--enable-gd-native-ttf' 
'--without-gdbm' '--with-gettext' '--with-gmp' '--with-iconv' '--with-jpeg-dir=/usr' '--with-openssl' '--with-png' '--with-pspell' 
'--with-expat-dir=/usr' '--with-pcre-regex=/usr' '--with-zlib' '--with-layout=GNU' '--enable-exif' '--enable-ftp' 
'--enable-magic-quotes' '--enable-sockets' '--enable-sysvsem' '--enable-sysvshm' '--enable-sysvmsg' '--enable-track-vars' 
'--enable-trans-sid' '--enable-yp' '--enable-wddx' '--with-kerberos' '--enable-ucd-snmp-hack' '--with-unixODBC=shared,/usr' 
'--enable-memory-limit' '--enable-shmop' '--enable-calendar' '--enable-dbx' '--enable-dio' '--without-mime-magic' 
'--without-sqlite' '--with-libxml-dir=/usr' '--with-xml' '--with-apxs2=/usr/sbin/apxs' '--without-mysql' '--without-gd' '--without-odbc' 
'--disable-dom' '--disable-dba' '--without-unixODBC' '--disable-pdo' '--disable-xmlreader' '--disable-xmlwriter' '--disable-json'

Server API 	Apache 2.0 Handler
Virtual Directory Support 	disabled
Configuration File (php.ini) Path 	/etc
Loaded Configuration File 	/etc/php.ini
Scan this dir for additional .ini files 	/etc/php.d
additional .ini files parsed 	/etc/php.d/json.ini, /etc/php.d/ldap.ini, /etc/php.d/mysql.ini, /etc/php.d/mysqli.ini, /etc/php.d/pdo.ini,
/etc/php.d/pdo_mysql.ini, /etc/php.d/pdo_sqlite.ini, /etc/php.d/zip.ini
PHP API 	20041225
PHP Extension 	20060613
Zend Extension 	220060519
Debug Build 	no
Thread Safety 	disabled
Zend Memory Manager 	enabled
IPv6 Support 	enabled
Registered PHP Streams 	php, file, data, http, ftp, compress.bzip2, compress.zlib, https, ftps, zip
Registered Stream Socket Transports 	tcp, udp, unix, udg, ssl, sslv3, sslv2, tls
Registered Stream Filters 	string.rot13, string.toupper, string.tolower, string.strip_tags, convert.*, consumed, convert.iconv.*, bzip2.*, zlib.*

    [SOLVED] could not connect to server: Permission denied

    in PostgreSQL

    Posted

    This is going to be an SELinux issue!  I thought I had SELinux turned off but I've just been trying to check and I think it's running in permissive mode.

     

     
[root@xyala ~]# /usr/sbin/sestatus | grep SELinux
SELinux status:                 enabled
SELinuxfs mount:                /selinux
[root@xyala ~]# /usr/sbin/sestatus | grep mode
Current mode:                   permissive

    I'm not sure that means it's in permissive mode but I can't get in front of the box to check what I set in the GUI.

     

    I've edited /etc/selinux/config to disable it fully but that needs a reboot to take effect and I can't reboot just at the moment.

     

    I'll let you know if it starts working (then again you'll probably hear me screaming about SELinux  >:( )

    [SOLVED] could not connect to server: Permission denied

    in PostgreSQL

    Posted

    How about compiling a command line installation of php in a local directory, and see if that works.

     

    Thanks, I never knew there was such a thing. I didn't compile a local version because I seem to already have it so tell me if I should have.

     

    I wrote this little bit of php

     

    <?php
$dbconn=pg_connect("dbname=lumbribase");
if ( ! $dbconn ) {
    echo "Error connecting to the database !<br> " ;
    printf("%s", pg_errormessage( $dbconn ) );
    exit(); }
else {echo "connected", "\n";}

$sqlcom="select * from lib";
$dbres = pg_exec($dbconn, $sqlcom );
if ( ! $dbres ) {
     echo "Error : " + pg_errormessage( $dbconn );
     exit();
}

$do = pg_Fetch_Object($dbres, 2);
$name=$do->name;
echo "and the name is... $name\n";
?>

     

    and ran it

    [ann@xyala tmp]$ php test.php 
connected
and the name is... Juvenile Earthworm Library

     

    (I only made it go and get "Juvenile Earthworm Library" because I didn't believe the "connected")

     

    ">which php" say's I'm using /usr/bin/php on the command line and I think the configure command details on here http://xyala.cap.ed.ac.uk/php_info.php means the web pages use the same php.

     

    Is there a difference in how the command line and web interfaces connect?

    [SOLVED] could not connect to server: Permission denied

    in PostgreSQL

    Posted

    Can anyone explain what going wrong here? I'm at a loss and the postgres mailing list hasn't come up with and answer.

    Any sugestions for a fix or how to debug this would be greatly appreciated.

     

    The message in the error_log is...

    PHP Warning:  pg_connect() [<a href='function.pg-connect'>function.pg-connect</a>]: Unable to connect to PostgreSQL server: could not connect to server: Permission denied\n\tIs the server running on host "localhost" and accepting\n\tTCP/IP connections on port 5432?

     

    The PHP code is...

    $dbconn=pg_connect( "dbname=lumbribase host=localhost port=5432 user=webuser" );
$dbconn=pg_connect(dbname=$PG_DATABASE);
if ( ! $dbconn ) {
    echo "Error connecting to the database !<br> " ;
    printf("%s", pg_errormessage( $dbconn ) );
    exit(); }

     

    According to the postgres guy's "psql -h localhost lumbribase"  means force postgres to use a TCP connection when making a local connection to postgres and that works fine.  Perl scripts using DBD_PG connect fine. Why is postgres denying permission to connections made through PHP? It's not user or database specific.  It would need to connect to the postgres server to check a users permissions but it's not getting that far.

     

    This bit's I know are...

    http://xyala.cap.ed.ac.uk/php_info.php say's php's configured for pgsql

     

    the port is accepting TCP

    [#@xyala]# telnet localhost 5432
Trying 127.0.0.1...
Connected to localhost.localdomain (127.0.0.1).
Escape character is '^]'.
Connection closed by foreign host.
[#@xyala]#

     

    pg_hba.conf is currently set up to allow anyone in

    [#@xyala]# less /var/lib/pgsql/data/pg_hba.conf
# TYPE  DATABASE    USER        CIDR-ADDRESS          METHOD

# "local" is for Unix domain socket connections only
local   all         all                               trust
host    all         all         127.0.0.1/32          trust
host    all         all         ::1/128               trust

     

    postgres.conf is set to listen on all ports

    [#@xyala]# grep 'listen' /var/lib/pgsql/data/postgresql.conf
# "pg_ctl reload". Some settings, such as listen_address, require
#listen_addresses = 'localhost' # what IP interface(s) to listen on;
listen_addresses = '*'
[#@xyala]#

     

    don't know what should be in this file but I can't think I've ever had to change it before.

    [#@xyala]# less /etc/php.d/pgsql.ini
; Enable pgsql extension module
extension=pgsql.so

     

    the server I'm going to replace is running the same versions of PHP and postgres http://zeldia.cap.ed.ac.uk/php_info.php

    The /etc/php.ini files on the two machines are the same and the /var/lib/pgsql/data/postgresql.conf files are only different because I've set listen_addresses = '*' on the new server (xyala) to see if I can make it work.

     

    Join the dots or rather the check boxes.

    in HTML Help

    Posted

    Hi

     

    Does anyone know of a way to get this <a href="http://trichuris.cap.ed.ac.uk/tmp/required.jpg">screenshot</a> type of display in all browsers.

     

    My curret set up (<a href="http://zeldia.cap.ed.ac.uk/clade_test.shtml">here</a>) uses a background image in a table and it's great in firefox if you don't change the size of the text but what good's that...

     

    So is there another way to tackle the problem; showing the relations between checkboxes (drawing the lines) relative to where the browser positions the checkboxes? Something like an image map but I don't think you can have checkboxes with them.

     

    Any advice/sugestions appreciated

    Thanks

    $_SERVER['PHP_SELF'] in php includes

    in HTML Help

    Posted

    Hi

    What I want is to be able to include a php file in an shtml doc and have a function in the php (like $_SERVER['PHP_SELF']) which will report the name of the shtml.

    e.g.
    if <b>test.shtml</b> calls species_db_queries.php like this
          <!--#include virtual="species_db_queries.php" -->

    then species_db_queries.php prints $_SERVER['PHP_SELF'] like ...
        $ref=$_SERVER['PHP_SELF'];
        print "$ref";

    I get "<b>species_db_queries.php</b>" printed when what I'm after is <b>test.shtml</b>.
    But similar code using test.php instead of test.shtnl gives me what I want. e.g.

    e.g. if <b>test.php</b> calls species_db_queries.php like this
        <? include ("species_db_queries.php"); ?>

    then species_db_queries.php prints $_SERVER['PHP_SELF'] like ...
        $ref=$_SERVER['PHP_SELF'];
        print "$ref";

    I get "<b>test.php</b>" printed.

    So can I get the required result and keep my shtml files as shtml?

    Thanks for your time
    Ann
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