I have big problem with my code. In loop while working only first argument if. Else doesn't work... Here is a code :
$sql="SELECT * FROM house1 WHERE week = '".$q."'";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result))
{
if ($row['house'] == $q)
{
echo '<img src="house2.gif"/>';
}
else
{
echo '<img src="house1.gif"/>';
}
}
Related
i receive "echo "Klaida!!!!"" meaning there is no results.
Could you please point out my mistake in code.
$dezesdezesId = $_GET["dezesId"];
$query = "SELECT * FROM dezes WHERE dezesId = '$dezesdezesId'";
$result = mysqli_query($connection, $query);
if (mysqli_num_rows ($result) > 0)
{
while ($row = mysqli_fetch_assoc($result))
{
$dezesPavadinimas = $row["pavadinimas"];
$dezesLikutis = $row["likutis"];
}
echo $dezesLikutis;
}
else
{
echo "Klaida!!!!" . mysqli_error($connection);
}
echo this inside your while loop.
while ($row = mysqli_fetch_assoc($result))
{
$dezesPavadinimas = $row["pavadinimas"];
$dezesLikutis = $row["likutis"];
echo $dezesLikutis;
}
and try to use '{$dezesdezesId}' your variable inside your query string
I want to access while array $row2 outside the loop so that I can compare in if else condition, because $row2 array contains many number. Is there any solution make $row array accessible outside the loop or any other method?
<?php
$sql = (" SELECT * FROM result ");
$query = mysqli_query($db, $sql);
while($row = mysqli_fetch_array($query)) {
$row2 $row['id'];
}
for($i=1;$i<=10;$i++) {
if ($i<= $row2) {
echo "<font color='red'><font size='50px'>".$i."</font></font>"."<br>";
continue;
} else {
echo $i.'<br>';
}
}
?>
If you want to loop over two dependent things you need to nest them or else $row2 will always contain the last value in the while loop:
<?php
$sql = (" SELECT * FROM result ");
$query = mysqli_query($db, $sql);
while ($row = mysqli_fetch_array($query)) {
$row2 = $row['id']; // I added an = sign here.
for($i=1;$i<=10;$i++) {
if ($i<= $row2) {
echo "<font color='red'><font size='50px'>".$i."</font></font>"."<br>";
continue;
} else {
echo $i.'<br>';
}
}
}
If this isn't what you want please clarify your question.
Pass your variable to a separate function that performs whatever task you want.
<?php
while($row = mysqli_fetch_array($query))
{
MyFunction($row2);
}
function MyFunction($passed_row2_value){
// perform whatever task you want here.
}
?>
This shows with some format the ids retrieved by the query and fills the blanks without format.
<?php
$sql = (" SELECT * FROM result ");
$query = mysqli_query($db, $sql);
$row2 = [];
while($row = mysqli_fetch_array($query)) {
$row2[$row['id']] = true;
}
for($i=1;$i<=10;$i++) {
if ($row2[$i]) {
echo "<font color='red'><font size='50px'>".$i."</font></font>"."<br>";
} else {
echo $i.'<br>';
}
}
?>
Any idea why this is showing a result of null? I'm guessing it has to do with where I put $link before I did the query, but it has to be done that way, correct?
function getcatposts($cat_id) {
$qc = #mysqli_query($link, "SELECT * FROM topics WHERE topic_cat='$cat_id'");
while($row = mysqli_fetch_array($qc)) {
$topic_title=$row['topic_subject'];
$topic_id=$row['topic_id'];
}
$qc2 = #mysqli_query($link, "SELECT * FROM categories WHERE cat_id='$cat_id'");
while($row2 = mysqli_fetch_array($qc2)) {
$cat_name=$row2['cat_name'];
}
Updated code:
function getcatposts($cat_id) {
$link = mysqli_connect("localhost", "lunar_lunar", "", "lunar_users");
$qc = mysqli_query($link, "SELECT * FROM topics WHERE topic_cat='$cat_id'");
while($row = mysqli_fetch_array($qc)) {
$topic_title=$row['topic_subject'];
$topic_id=$row['topic_id'];
}
$qc2 = mysqli_query($link, "SELECT * FROM categories WHERE cat_id='$cat_id'");
while($row2 = mysqli_fetch_array($qc2)) {
$cat_name=$row2['cat_name'];
}
echo $cat_name;
echo '<br />';
echo $topic_title;
echo '<br />';
echo $topic_id;
}
New issue is that its displaying like this:
http://gyazo.com/43e8a91b9e0cf4f5e413536907891dcf.png
When the DB looks like this:
http://gyazo.com/1ead8bd0f150838dae3ee4a476419679.png
It should be displaying all three of them and this is a function meaning it will keep redoing all the code until it can't query anymore data. Any ideas?
The problem here is that you're trying to echo the values outside your loop. The variables inside the loop will get overwritten on each iteration and at the end of looping, the variable will hold the value of the last iteration.
If you want to display all the values, move the echo statement inside your loop, like so:
while($row = mysqli_fetch_array($qc))
{
$topic_title = $row['topic_subject'];
$topic_id = $row['topic_id'];
echo $topic_title.'<br/>';
echo $topic_id.'<br/>';
}
$qc2 = mysqli_query($link, "SELECT * FROM categories WHERE cat_id='$cat_id'");
while($row2 = mysqli_fetch_array($qc2))
{
$cat_name = $row2['cat_name'];
echo $cat_name.'<br/>';
}
If you care about the order, you could store the titles, ids and cat_names in arrays like so:
while($row = mysqli_fetch_array($qc))
{
$topic_title[] =$row['topic_subject'];
$topic_id[] = $row['topic_id'];
}
$qc2 = mysqli_query($link, "SELECT * FROM categories WHERE cat_id='$cat_id'");
while($row2 = mysqli_fetch_array($qc2))
{
$cat_name[] =$row2['cat_name'];
}
And then loop through them:
for ($i=0; $i < count($topic_id); $i++) {
if( isset($topic_id[$i], $topic_title[$i], $cat_name[$i]) )
{
echo $cat_name[$i].'<br/>';
echo $topic_title[$i].'<br/>';
echo $topic_id[$i].'<br/>';
}
}
Your function displays only one result because your echo is outside the while loop...
Put the Echo statements inside the loop, or you will print only the last result!
while($row = mysqli_fetch_array($qc)) {
$topic_title=$row['topic_subject'];
$topic_id=$row['topic_id'];
echo $topic_title;
echo '<br />';
echo $topic_id;
}
$qc2 = mysqli_query($link, "SELECT * FROM categories WHERE cat_id='$cat_id'");
while($row2 = mysqli_fetch_array($qc2)) {
$cat_name=$row2['cat_name'];
echo $cat_name;
echo '<br />';
}
the code i have below basically behaves as a friend search engine where you can add someone if you don't have them on you account. So i am trying to do the following: If exists it should just display the friend and if not then it should say add friend.
I got it to work but the tail end of the query (please see this comment "//Problem code - when i use this echo below this where it repeats 3 times") is giving me trouble where it repeats Add 3 times.
<?php
$num_rows1 = mysql_num_rows($result);
if ($result == "") {
echo "";
}
echo "";
$rows = mysql_num_rows($result);
if ($rows == 0) {
print("<div id=norequests>No results for <strong>$q
</strong></div>");
}
elseif ($rows > 0) {
while ($row = mysql_fetch_array($query))
{
$person = htmlspecialchars($row['full_name']);
$linksys = htmlspecialchars($row['name']);
$pid = htmlspecialchars($row['system_id']);
}
print("");
}
}
else{
echo '<div id="error">No results.</div>';
}
$sql = "SELECT `Friend_id` from `friends_container`
WHERE `System_id` = '$sid'";
$result = mysql_query($sql);
$query = mysql_query($sql) or die("Error: " . mysql_error());
if ($result == "") {
echo "";
}
echo "";
$rows = mysql_num_rows($result);
if ($rows == 0) {
print("");
}
elseif ($rows > 0)
{
while ($row = mysql_fetch_array($query))
{
$existing = htmlspecialchars($row['Friend_id']);
if ($existing == $pid) {
echo("<img src=$linksys />$person - Already Existing");
}
else
//Problem code - when i use this echo below this where it repeats 3 times
{
echo("Add $person");
}
}
?>
You must have 3 images stored for each account.
Your query will always result in a multiple result. what you should do is use php to convert it to something which will result in a better option or say convert the result in an array for convenience.
I Have a PHP file doing a query
function getFiles()
{
$solnid = $_GET[id];
$query = mysql_query("SELECT * FROM library WHERE libr_solutionId = '$solnid' AND libr_deleted IS NULL") or die(mysql_error());
$result = mysql_query($query);
if (mysql_num_rows($result)==0) {
echo "<br>No Related Links";
} else {
while($library = mysql_fetch_assoc($query)) {
echo "<span style=\"margin-bottom: 5px;\"><img src=\"images/icon-download-02.png\" align=\"absmiddle\" style=\"margin-right: 5px;\"><span>".$library[libr_c_title]."</span></span><br>";
echo "<br>";
}
}
}
I have tested the query in PHPMyAdmin and I get a result however when running the function in my page it only displays the 'No Related Links' statement any ideas anyone?
Here's the fixed code. :) The problem was that you used mysql_query on a Resource (the first mysql_query), and that will give 0 rows, or an exception..
function getFiles()
{
$solnid = $_GET['id'];
$result = mysql_query("SELECT * FROM library WHERE libr_solutionId = '$solnid' AND libr_deleted IS NULL") or die(mysql_error());
if (mysql_num_rows($result)==0) {
echo "<br>No Related Links";
} else {
while($library = mysql_fetch_assoc($result)) {
echo "<span style=\"margin-bottom: 5px;\"><img src=\"images/icon-download-02.png\" align=\"absmiddle\" style=\"margin-right: 5px;\"><span>".$library['libr_c_title']."</span></span><br>";
echo "<br>";
}
}
}
change your function to (change to lines)
function getFiles()
{
$solnid = $_GET[id];
$result = mysql_query("SELECT * FROM library WHERE libr_solutionId = '$solnid' AND libr_deleted IS NULL") or die(mysql_error());
if (mysql_num_rows($result)==0) {
echo "<br>No Related Links";
} else {
while($library = mysql_fetch_assoc($query)) {
echo "<span style=\"margin-bottom: 5px;\"><img src=\"images/icon-download-02.png\" align=\"absmiddle\" style=\"margin-right: 5px;\"><span>".$library[libr_c_title]."</span></span><br>";
echo "<br>";
}
}
}