I have the code below and it's working as it should however instead of echoing the results to the screen, I need to store the results in a php variable called $var. How would I go about doing that?
<?php
$sql = "SELECT id_member FROM smf_members WHERE FIND_IN_SET(24,additional_groups)";
$con = mysql_connect('localhost', 'sqluser', 'sqlpass');
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("test_db");
$result = mysql_query开发者_运维百科($sql, $con);
while ($row = mysql_fetch_array($result)) {
echo $row['id_member'];
}
mysql_close($con);
?>
Depending on what you want to achieve here is a few possible ways of doing this
$var = "";
while ($row = mysql_fetch_array($result)) {
$var .= $row['id_member'] . "\n";
}
$var = array();
while ($row = mysql_fetch_array($result)) {
$var[] = $row['id_member'];
}
replace echo
with $var[]
.
That will push each result onto the end of the array. It would probably be good to define the variable first.
$var = array();
while ($row = mysql_fetch_array($result)) {
$var[] = $row['id_member'];
}
<?php
$sql = "SELECT id_member FROM smf_members WHERE FIND_IN_SET(24,additional_groups)";
$con = mysql_connect('localhost', 'sqluser', 'sqlpass');
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("test_db");
$result = mysql_query($sql, $con);
$v = array(); // $v instead of $var, since var is a keyword and may cause troubles
while ($row = mysql_fetch_array($result)) {
array_push($v, $row['id_member']);
// or
//$v[] = $row['id_member'];
}
mysql_close($con);
?>
If the select statement will return more than one result, then you need to store the values in an array:
$member_ids = array();
while ($row = mysql_fetch_array($result)) {
$member_ids[] = $row['id_member'];
}
If the select statement will return a single result (you can make sure by appending a LIMIT 1
to the value of the $sql
variable).
$row = mysql_fetch_array($result);
$member_id = $row['id_member'];
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