Jiggens
asked on
If Statement for sepecifc link to PDF
I am still new to PHP and was wondering how i would put in a specifc link to a pdf file on my page to show up under one set of data but not the other.
<?
if (mysql_fetch_row(mysql_que
print "<p>Coming Soon</p><p> </p>";
} else {
print "<table border='0' cellspacing='0' cellpadding='5'>";
while($row = mysql_fetch_row($result)) {
print "<tr><td valign='top' align='center'><a href='community.php?ID=" . $row[0] . "'><img src='../images/browse/logo
print "<td valign='top'><p><span class='comm-header-excepti
print $row[5] . "<br />";
print "<a href='community.php?ID=" . $row[0] . "'>View Details</a></p>";
print "</td></tr>";
}
print "</table>";
}
?>
Under view details i want to put another link for Self-Qual Form and the location is http://homes.pacificscene.com/browse/selfqualform/ and i just want to add it as hyperlink for community.php?ID=43 and no others.
Could someone give me a hand on how to do this?
<?
if (mysql_fetch_row(mysql_que
print "<p>Coming Soon</p><p> </p>";
} else {
print "<table border='0' cellspacing='0' cellpadding='5'>";
while($row = mysql_fetch_row($result)) {
print "<tr><td valign='top' align='center'><a href='community.php?ID=" . $row[0] . "'><img src='../images/browse/logo
print "<td valign='top'><p><span class='comm-header-excepti
print $row[5] . "<br />";
print "<a href='community.php?ID=" . $row[0] . "'>View Details</a></p>";
print "</td></tr>";
}
print "</table>";
}
?>
Under view details i want to put another link for Self-Qual Form and the location is http://homes.pacificscene.com/browse/selfqualform/ and i just want to add it as hyperlink for community.php?ID=43 and no others.
Could someone give me a hand on how to do this?
ASKER CERTIFIED SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Jiggens,
What I would prescribe is that you loop through your dataset and perform a test condition inside that loop. See below:
// Prepare your query
$qry = "SELECT * from psh_communities WHERE status='Current' and area='San Diego'";
// Perform query
$result = mysql_query($qry) or die("this query is no good, here's why: " . mysql_error());
// Loop through query result
while ($row = mysql_fetch_array($result)
if($row['_CRITERIA_'] != TRUE){
print "<p>Coming Soon</p><p> </p>";
} else {
print "<table border='0' cellspacing='0' cellpadding='5'>";
while($row = mysql_fetch_row($result)) {
print "<tr><td valign='top' align='center'><a href='community.php?ID=" . $row[0] . "'><img src='../images/browse/logo
print "<td valign='top'><p><span class='comm-header-excepti
print $row[5] . "<br />";
print "<a href='community.php?ID=" . $row[0] . "'>View Details</a></p>";
print "</td></tr>";
}
print "</table>";
}
where _CRITERIA_ is whatever piece of information that your data has that you need to test against.
Does this help?
ps, as a long time PHP developer, I suggest you reference your return data by name instead of index. mysql_fetch_array() will generate both index and a name reference per column in each row. so:
TABLE
ID NAME PHONE
1 chris 12345
2 bob 67890
$res = mysql_query('select * from TABLE');
mysql_fetch_array($res) will return something like array([0]=>1, ['ID']=>1, [1] => 'chris', ['NAME'] => 'chris', [2]=>'12345', ['PHONE']=>'12345'). Does this make sense?
What I would prescribe is that you loop through your dataset and perform a test condition inside that loop. See below:
// Prepare your query
$qry = "SELECT * from psh_communities WHERE status='Current' and area='San Diego'";
// Perform query
$result = mysql_query($qry) or die("this query is no good, here's why: " . mysql_error());
// Loop through query result
while ($row = mysql_fetch_array($result)
if($row['_CRITERIA_'] != TRUE){
print "<p>Coming Soon</p><p> </p>";
} else {
print "<table border='0' cellspacing='0' cellpadding='5'>";
while($row = mysql_fetch_row($result)) {
print "<tr><td valign='top' align='center'><a href='community.php?ID=" . $row[0] . "'><img src='../images/browse/logo
print "<td valign='top'><p><span class='comm-header-excepti
print $row[5] . "<br />";
print "<a href='community.php?ID=" . $row[0] . "'>View Details</a></p>";
print "</td></tr>";
}
print "</table>";
}
where _CRITERIA_ is whatever piece of information that your data has that you need to test against.
Does this help?
ps, as a long time PHP developer, I suggest you reference your return data by name instead of index. mysql_fetch_array() will generate both index and a name reference per column in each row. so:
TABLE
ID NAME PHONE
1 chris 12345
2 bob 67890
$res = mysql_query('select * from TABLE');
mysql_fetch_array($res) will return something like array([0]=>1, ['ID']=>1, [1] => 'chris', ['NAME'] => 'chris', [2]=>'12345', ['PHONE']=>'12345'). Does this make sense?
Jiggens,
i don't think I correctly answered your question. Also I didn't really review your code beyond the if / else statement, and so what I posed doesn't really make sense.
What I think you want to do is output a set of data, but only show a particular link for some parts of that data. Is that correct?
i don't think I correctly answered your question. Also I didn't really review your code beyond the if / else statement, and so what I posed doesn't really make sense.
What I think you want to do is output a set of data, but only show a particular link for some parts of that data. Is that correct?
Jiggens,
I'm curious what you mean by "I want to print it if the link is ID=43", do you mean you want it on the form if that is the requesting URL, or you want it on the form if the generated link is ID=43? if it's when the gentrated link is ID=43 then TerryAtOpus would appear to have the correct answer, and the code is already in the right place.
If the link needs to be put there when the requesting ID=43 in the query string then you'll need to do the following:
print "<a href='community.php?ID=" . $row[0] . "'>View Details</a><br/>";
if( $_GET['ID'] == '43' )
{
print " <a href='http://homes.pacificscene.com/browse/selfqualform/'>Self-Qual Form</a>";
}
print "</p>";
print "</td></tr>";
I'm curious what you mean by "I want to print it if the link is ID=43", do you mean you want it on the form if that is the requesting URL, or you want it on the form if the generated link is ID=43? if it's when the gentrated link is ID=43 then TerryAtOpus would appear to have the correct answer, and the code is already in the right place.
If the link needs to be put there when the requesting ID=43 in the query string then you'll need to do the following:
print "<a href='community.php?ID=" . $row[0] . "'>View Details</a><br/>";
if( $_GET['ID'] == '43' )
{
print " <a href='http://homes.pacificscene.com/browse/selfqualform/'>Self-Qual Form</a>";
}
print "</p>";
print "</td></tr>";
print "<a href='test.pdf'>PDF Link</a>";
}
Place this wherever you want the link.