[SOLVED] SHOW TABLES only getting 1st table...

RIRedinPA · October 21, 2009

I got a db with about 40 tables, I want to list these in a drop down for users. I'm getting the correct number of results but I am parsing the data wrong and only getting the first table. Not sure what I am doing wrong here.


        $query = "SHOW TABLES"; 
$result = mysql_query($query);
$a=0; 
while($showtablerow = mysql_fetch_array($result)) {
	$tablename = $showtablerow[$a];
	//$tablenamearray = explode("_", $tablename); 
	$emparray[] =  $tablename;
	$a++; 
}

I use $emparray to then build the SELECT html...

cags · October 21, 2009

$query = "SHOW TABLES";
$result = mysql_query($query);
$emparray = array();

while($showtablerow = mysql_fetch_array($result)) {
   $emparray[] = $showtablerow[0];
}

RIRedinPA · October 22, 2009

$query = "SHOW TABLES";
$result = mysql_query($query);
$emparray = array();

while($showtablerow = mysql_fetch_array($result)) {
   $emparray[] = $showtablerow[0];
}

Thanks Cags but I had previously tried $showtablerow[0] and got the same result. I get 40 results but all with the 1st table name.

cags · October 22, 2009

Are you saying you have tried that block of code and it's not working?

RIRedinPA · October 22, 2009

Are you saying you have tried that block of code and it's not working?

I had something similar before and it didn't work. I plugged your code in and it didn't work either, same results, I'm only getting the first table name.

The tables are named by employee last name and ISO Week # so they look like this:

Jones_0751

Jones_0752

Jones_0801

Jones_0802

...

Jones_0943

Smith_0751

Smith_0752

...

and so on...

I get the right number of results but only the first table name - so my result list looks like this:

Jones_0751

Jones_0751...

RIRedinPA · October 22, 2009

Here's the full code I am now using:

include "config.php";

$thistask = $_GET['thistask'];

if ($thistask == "show1") { 

$emparray = Array();

//get employee list
$tableselect = "<fieldset><select name=\"activitysheets\" onchange=\"loadissuevalues(this.value, '$magcode');\"><option value=\"null\">Select An Issue Date</option> ";
$emparray = array();

$query = 'SHOW TABLES FROM RecAdDept';
$result = mysql_query($query) or die ("Cannot get tables\n\n" . mysql_error());
while($table = mysql_fetch_row($result)) {
	$emparray[] = $table[0]; 
} 


//clean array of duplicates
//$emparray = array_unique($emparray);

//loop through to add options to select html
for ($x=0; $x<count($emparray); $x++) { 
	$tableselect .= "<option value='$emparray[0]'>$emparray[0]</option>";
}

$tableselect .= "</select></fieldset>";

echo $tableselect;

} else { 
echo "show all tables - ug.";
}

Here's my return:

<option value="null">Select An Issue Date</option>

<option value='Jones_0751'>Jones_0751</option>

...

<option value='Jones_0751'>Jones_0751</option>

</select>

</fieldset>

RIRedinPA · October 22, 2009

Here's the whole setup:

Page where the user selects which task they want to perform. I'm not doing anything with the include files yet so they have no affect on output.:


<!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>Merion Recruitment Ad Department Activity Sheet Manager</title>
<link href="css/radas.css" rel="stylesheet" type="text/css" />

<script language="javascript" src="js/radas.js"></script>
</head>

<body onload="setup();">

<?php 
include ('lib/config.php');
include ('lib/functions.php');
include ('lib/arrays.php');
?>

<div id="header"><a href="index.php?view=new"><img src="images/title.gif" border="0"></a></div>

<div id="leftCol">

    Select a task from the options below:
    <p>
    
    <form name="getTask" method=GET action="index.php">
    <fieldset>
    	<legend>Admin Options</legend>
   	 	<select name="taskChoice" onchange="showemployees()">
    		<option value="null">Select A Task</option>
        	<option value="view1">View Employee</option>
        	<option value="view2">View All Employees</option>
    	</select>
    </fieldset>
    
    <div id="emptable" style="display: none;">
    	 
    </div>
    </form>
</div>

<div id="resultTable"> </div>

</body>
</html>

Javascript to get data, make AJAX call:


// JavaScript Document

function showemployees() { 


//set form
var theform = document.getTask; 

//determine if one or all employees
var taskChoiceSI = theform.taskChoice.selectedIndex;
if (taskChoiceSI != 0) { 
	if (taskChoiceSI == 1) { 
		var thistask = "show1";
	} else { 
		var thistask = "showall";
	}

	var xmlHttp = checkajax();

	xmlHttp.onreadystatechange=function() {
		if(xmlHttp.readyState==4) {
			alert(xmlHttp.responseText);
			//document.getElementById('emptable').innerHTML = xmlHttp.responseText;
			//document.getElementById('emptable').style.display = "block";
		}
	}

	xmlHttp.open("GET", "lib/getemplist.php?thistask=" + thistask, true);
	xmlHttp.send(null);
}



}

function setup() { 
document.getTask.taskChoice[0].selected = true; 
}


function checkajax() {
//ajax compatibility check
	var xmlHttp;
	try {
  			// Firefox, Opera 8.0+, Safari
  			xmlHttp=new XMLHttpRequest();
  		}
		catch (e) {
  			// Internet Explorer
  			try {
    			xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
    		}
  				catch (e) {
    			try {
      				xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
      			}
    				catch (e) {
      				alert("Your browser does not support AJAX!");
      				return false;
      			}
    		}
  		}

	return xmlHttp
}

and the gettemplist.php code, which I also posted above:


<?php 


include "config.php";

$thistask = $_GET['thistask'];


if ($thistask == "show1") { 

$emparray = Array();


//get employee list
$tableselect = "<fieldset><select name=\"activitysheets\" onchange=\"loadissuevalues(this.value);\"><option value=\"null\">Select An Issue Date</option> ";
$emparray = array();

$query = 'SHOW TABLES FROM RecAdDept';
$result = mysql_query($query) or die ("Cannot get tables\n\n" . mysql_error());
while($table = mysql_fetch_row($result)) {
	$emparray[] = $table[0]; 
} 


//clean array of duplicates
//$emparray = array_unique($emparray);

//loop through to add options to select html
for ($x=0; $x<count($emparray); $x++) { 
	$tableselect .= "<option value='$emparray[0]'>$emparray[0]</option>";
}

$tableselect .= "</select></fieldset>";

echo $tableselect;

} else { 
echo "show all table - ug.";
}
?>

DavidAM · October 22, 2009

   //loop through to add options to select html
   for ($x=0; $x<count($emparray); $x++) {
      $tableselect .= "<option value='$emparray[0]'>$emparray[0]</option>";
   }

You have collected the table names in an array ($emparray). In the loop, you are always referring to the first element of the array ($emparray[0]), so you are just getting the first table. Change that to:

   //loop through to add options to select html
   for ($x=0; $x<count($emparray); $x++) {
      $tableselect .= "<option value='$emparray[$x]'>$emparray[$x]</option>";
   }

cags · October 22, 2009

The reason your output is wrong has nothing to do with the original set of code, the code I gave before works perfectly.

for ($x=0; $x<count($emparray); $x++) {
   $tableselect .= "<option value='$emparray[0]'>$emparray[0]</option>";
}

You're always echo'ing out the first item in the array, the 0 should be $x to print out the current item. Having said that the whole content of the if block may just as well be...

$tableselect = "<fieldset><select name=\"activitysheets\" onchange=\"loadissuevalues(this.value, '$magcode');\"><option value=\"null\">Select An Issue Date</option> ";   
$query = 'SHOW TABLES FROM RecAdDept';
$result = mysql_query($query) or die ("Cannot get tables\n\n" . mysql_error());
while($table = mysql_fetch_row($result)) {
$tableselect .= "<option value='".$table[0]."'>".$table[0]."</option>";
}    
$tableselect .= "</select></fieldset>"; 
echo $tableselect;

Eliminating the need for 2 loops.

RIRedinPA · October 22, 2009

The reason your output is wrong has nothing to do with the original set of code, the code I gave before works perfectly.
for ($x=0; $x<count($emparray); $x++) {
   $tableselect .= "<option value='$emparray[0]'>$emparray[0]</option>";
}
You're always echo'ing out the first item in the array, the 0 should be $x to print out the current item. Having said that the whole content of the if block may just as well be...
$tableselect = "<fieldset><select name=\"activitysheets\" onchange=\"loadissuevalues(this.value, '$magcode');\"><option value=\"null\">Select An Issue Date</option> ";   
$query = 'SHOW TABLES FROM RecAdDept';
$result = mysql_query($query) or die ("Cannot get tables\n\n" . mysql_error());
while($table = mysql_fetch_row($result)) {
$tableselect .= "<option value='".$table[0]."'>".$table[0]."</option>";
}    
$tableselect .= "</select></fieldset>"; 
echo $tableselect;
Eliminating the need for 2 loops.

Damn, that was embarrassing... thanks for the help...

cags · October 22, 2009

No problem, your welcome. There is a SOLVED button at the bottom left corner of your screen so people know you've solved the problem.

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[SOLVED] SHOW TABLES only getting 1st table...

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