WHERE statement = mySQL value

dfalkowitz · July 28, 2011

Is there a way to echo a value from a mySQL database? For example:

table

ID AGE

1 21

2 22

PHP code

$query = "SELECT * FROM table WHERE 'ID' = ?????????";

$result = mysql_query($query) or die(mysql_error());

$row = mysql_fetch_array($result) or die(mysql_error());

echo $row['AGE'];

What would go in replace of the "????????"

dfalkowitz · July 28, 2011

One thing to add - I want to return only the AGE of either ID = 1 or ID = 2. I have a form drop-down with option 1 and 2, if option 1 is chosen, I only want to show the AGE for ID = 1...

WebStyles · July 28, 2011

$q = mysql_query("SELECT * FROM table WHERE `ID` = '1'") or die(mysql_error());
$r = mysql_fetch_assoc($result) or die(mysql_error());
echo $r['AGE'];

Muddy_Funster · July 28, 2011

Why have you put backticks around ID? and why have you put the number 1 inside quotes?

WebStyles · July 28, 2011

personal habit.

I always use backticks around field names, and always enclose the values in single quotes.

dfalkowitz · July 28, 2011

Thanks for replies!

I don't want this because this will only show the AGE for option 1

$q = mysql_query("SELECT * FROM table WHERE `ID` = '1'") or die(mysql_error());

$r = mysql_fetch_assoc($result) or die(mysql_error());

echo $r['AGE'];

I need ID to = a variable. If option 1 is chosen then show AGE for ID = 1 but if option 2 is chosen then show AGE for ID = 2...

WebStyles · July 28, 2011

replace 1 with variable.

$q = mysql_query("SELECT * FROM table WHERE `ID` = '$variableName'") or die(mysql_error());
$r = mysql_fetch_assoc($result) or die(mysql_error());
echo $r['AGE'];

Muddy_Funster · July 28, 2011

personal habit.

I always use backticks around field names, and always enclose the values in single quotes.

Always using quotes around values will meen that you can not ever use integer values then as wraping a value within quotes in SQL tells it it's a string.

WebStyles · July 28, 2011

really? so if ID field in database is an int, you think

SELECT * from `tableName` where `id` = '1'

doesn't work?

dfalkowitz · July 28, 2011

What would the variable be?

$row['ID'] or $_GET['ID']

WebStyles · July 28, 2011

the variable depends on how you're passing it over from the previous page:

$variableName = $_GET['variableName'];

$variableName = $_POST['variableName'];

WebStyles · July 28, 2011

relative to the personal message you sent me with your code:

at a (very) quick glance, your form does not have an action or method.

instead of just

<form>

use something like:

<form action="pageThatProcessesForm.php" method="get">

method can also be "post".

dfalkowitz · July 28, 2011

I still get same errror:

Notice: Undefined index: id in C:\xampp\htdocs\php_test\test.php on line 11 (which is $variableName = $_POST['ID']

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\php_test\test.php on line 14 (which is - while ($row = mysql_fetch_array($result)))

premiso · July 28, 2011

dfalkowitz, post your form HTML code. As we cannot be of further assistance without seeing how that is setup.

WebStyles · July 28, 2011

premiso, he sent it to me in a personal message (don't know why)

here it is:

<html>
<head>
<script type="text/javascript">
function showUser(str)
{
if (str=="")
{
document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
   {
   document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
   }
}
xmlhttp.open("GET","test.php?q="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>

<form>
<select name="users">
<option value="">Select ID:</option>
<option value="1">1</option>
<option value="2">2</option>
</select>
<input name="test" type="button" value="submit" onClick="showUser(this.value)">
<input name="test" type="text" id="txthint" size="50">
</form>
<br />
<div id="txtHint"><b>Age info will be listed here.</b></div>

</body>
</html>

dfalkowitz · July 28, 2011

test.php below

<?php

$dbhost = "localhost";

$dbuser = "test";

$dbpass = "test";

$dbname = "php_test";

//Connect to MySQL Server

mysql_connect($dbhost, $dbuser, $dbpass);

//Select Database

mysql_select_db($dbname) or die(mysql_error());

$variableName = $_GET['ID'];

$result = mysql_query("SELECT * FROM test WHERE id = $variableName");

while ($row = mysql_fetch_array($result))

{

echo $row['AGE'];

echo "<br />";

}

?>

premiso · July 28, 2011

$variableName = $_GET['ID'];

This should correspond with the name of your "users" select.

$variableName = $_GET['users'];

Since you named it 'users' you use users. If you want it to be ID, change the select box name to be ID.

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WHERE statement = mySQL value

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