how do I traverse rows in a table - php

I have a table named members with 3 rows. The following code attempts to display all the rows in the members table. It displays the 1'st record 3 times instead of displaying each record once.
<?php
$con = new mysqli("localhost", "root", "jce123", "profile");
if (mysqli_connect_errno()) {
printf("Connection failed: %s\n", mysqli_connect_error());
exit();
}
$sql = "SELECT * FROM members";
$res = mysqli_query($con,$sql);
$num = mysqli_num_rows($res);
$array = mysqli_fetch_assoc($res);
$keysarray = array_keys($array);
$valuearray = array_values($array);
for ($i=0; $i < $num; $i++) {
for($j = 0; $j < count($array); $j++) {
echo "<table>";
echo "<tr><td>".$keysarray[$j].": </td><td>".$valuearray[$j]."</td></tr>";
echo "</table>";
}
echo "<br><br>";
}
mysqli_close($con);
?>
I've updated this per suggestions:
$sql = "SELECT * FROM members";
$res = mysqli_query($con,$sql);
$num = mysqli_num_rows($res);
$array = mysqli_fetch_assoc($res);
while ($row = mysqli_fetch_assoc($res)) {
foreach($array as $key => $value){
echo "<table>";
echo "<tr><td>".$key.": </td><td>".$value."</td></tr>";
echo "</table>";
}
echo "<br><br>";
}

That's because, your $num = 3, count($array) = 3 and the outer for loop has no bearing on inner for loop.

Where you have $array = mysqli_fetch_assoc($res);, you will only receive one row from your database. You need something like:
while ($row=mysqli_fetch_assoc($res))
{
// processing for each row
}

Any basic tutorial will tell you this. Even the PHP docs provide an example.

you can use do something like this..
while ($row=mysqli_fetch_assoc($res))
{
// processing for each row
e.g
echo $row['id'];
echo $row['name'];
etc
}

try this... this peace of code is not tested but just to give you an idea...
$sql = "SELECT * FROM members";
$res = mysqli_query($con,$sql);
$num = mysqli_num_rows($res);
while ($row = mysqli_fetch_assoc($res)) {
echo "<table>";
echo '<tr><td> id = "'.$row['id'].'":
</td><td> name = "'.$row['name'].'"</td></tr>";
echo "</table>";
}

Related

How to store table Field(column) name into An Array?

This is the my code and I am trying to store the value into the Array.
$result = $cid->query("SHOW COLUMNS FROM commissions");
echo $count = $result->num_rows;
while ($row = $result->fetch_assoc()) {
echo $row['Field'];
echo "<br/>";
}
I am trying to do this $ar[]=$row; but there is nothing display !!
$result = $cid->query("SHOW COLUMNS FROM commissions");
echo $count = $result->num_rows;
while ($row = $result->fetch_assoc()) {
ar[] = $row;
echo "<br/>";
}
for($i = 0, $i<12, $i++) {
echo ar['field'][0];
}
Nothing Display in the output
You can do following way to store column name into array. Read comment after every line
For PDO
$rs = $cid->query('SELECT * FROM commissions');// your query
for ($i = 0; $i < $rs->field_count(); $i++) {/// count nu of column
$col = $rs->fetch_fields()($i);//Returns metadata for a column in a result set
$columns[] = $col['name'];// get name from metedata
}
print_r($columns);
For mysqli
$rs = $cid->query('SELECT * FROM commissions');// your query
$col = $rs->fetch_fields();//Returns metadata for a column in a result set
foreach ($col as $val) {
$columns[]=$val->name;
}
print_r($columns);

Check if FK order is numbered correctly

I have some absences, and each absence has an FK to an employee, which I wanna display in a generated HTML table. However, if I delete an employee the order changes from 0, 1, 2 to 1, 2 (for example if the second employee got deleted).
This messed up my code, because I need the FK to check where I have to insert the leave (in which <tr>). Here I count the employees:
$result = mysql_query("select count(1) FROM employee");
$row = mysql_fetch_array($result);
$count_user = $row[0];
Later in my code I make a query in a loop. The loop runs as many times as there are users. And heres the problem: if an employee gets deleted it will not reach the other one ($i goes 0 and then 1 if there are 2 users, but if one is deleted the FK is 3, so it would need to go one further).
to
$result = mysql_query("select start, end, type_FK, employee_FK FROM absences where employee_FK = $i");
while ($row = mysql_fetch_assoc($result)) {
$array_absences[] = $row;
}
Has anyone an Idea that is not based on the FK order?
Also this topic needs a better title so feel free to edit it.
<html>
<head>
<title>Absence System</title>
</head>
<body>
<?php
$con = mysql_connect("localhost", "root", "");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("absence_system", $con);
$result = mysql_query("select count(1) FROM employee");
$row = mysql_fetch_array($result);
$count_user = $row[0];
$result = mysql_query("select count(1) FROM absences");
$row = mysql_fetch_array($result);
$count_absences = $row[0];
$result = mysql_query("select name, surename, employee_ID FROM employee");
while ($row = mysql_fetch_assoc($result)) {
$array_user[] = $row;
}
for($i = 0; $i < $count_user; $i++){
echo '<table = border = 1px>';
echo '</tr>';
$result = mysql_query("select start, end, type_FK, employee_FK FROM absences where employee_FK = $i");
while ($row = mysql_fetch_assoc($result)) {
$array_absences[] = $row;
}
$count = count($array_absences);
echo $count;
print_r($array_absences);
for($j = 0; $j < 32; $j++){
$true = 0;
if($j == 0){
echo '<td>';
echo $array_user[$i]['name'], " ", $array_user[$i]['surename'] ;
echo '</td>';
}
for($k = 0; $k < $count; $k++)
{
$array_absences[$k]['start'] = substr($array_absences[$k]['start'], -2);
$array_absences[$k]['end'] = substr($array_absences[$k]['end'], -2);
$array_absences[$k]['start'] = ereg_replace("^0", "", $array_absences[$k]['start']);
$array_absences[$k]['end'] = ereg_replace("^0", "", $array_absences[$k]['end']);
if($j == $array_absences[$k]['start'] && $array_absences[$k]['employee_FK'] == $i){
$true = 1;
echo '<td>';
echo $array_absences[$k]['type_FK'];
echo '</td>';
}
}
if($j != 0 && $true == 0){
echo '<td>';
echo "$j";
echo '</td>';
}
}
echo '</tr>';
echo '</table>';
}
?>
</body>
</html>
If employee_ID corresponds to employee_FK, then you should not iterate over $count_user, but rather over $array_user - then your select would looke like this:
$result = mysql_query("select start, end, type_FK, employee_FK FROM absences where employee_FK = {$array_user[$i]['employee_ID']}");
while $i will still represent sequence number, because
$count_user == count($array_user)

Displaying data horizontally using php and mysql

Hi I'm attempting to display data retrieved from a mysql table horizontally in an html table using php. The code below works well except for the fact that it leaves out the first record (starts at the second record) in my database. I'm sure it has something to do with the counter but I can't seem to figure out how to get it to stop doing this. If anyone can point out my error I'd really appreciate it!
$items = 5;
$query = "SELECT * FROM members ";
$result = mysql_query($query)
or die(mysql_error());
$row = mysql_fetch_array($result);
if (mysql_num_rows($result) > 0) {
echo '<table border="1">';
$i = 0;
while($row = mysql_fetch_array($result)){
$first_name = $row['first_name'];
if ($i==0) {
echo "<tr>\n";
}
echo "\t<td align=\center\">$first_name</td>\n";
$i++;
if ($i == $items) {
echo "</tr>\n";
$i = 0;
}
}//end while loop
if ($i > 0) {
for (;$i < $items; $i++) {
echo "<td> </td>\n";
}
echo '</tr>';
}//end ($i>0) if
echo '</table>';
}else {
echo 'no records found';
}
try and remove the 1st
$row = mysql_fetch_array($result);
you are calling it twice, that's why it skips 1 row in your while loop
try this simpler.
$items = 5;
$query = "SELECT * FROM members ";
$result = mysql_query($query) or die(mysql_error());
if (mysql_num_rows($result) > 0) {
echo '<table border="1">';
while($row = mysql_fetch_array($result)){
$first_name = $row['first_name'];
echo "<tr>";
for ($i=0 ; $i <= $items ;$i++) {
echo "<td align='center'>".$first_name."</td>";
}
}//end while loop
echo "</tr>";
echo '</table>';
}else{ echo 'no records found'; }
I have run into this issue before. Try the do while loop instead. Example
do {
// code
} while($row = mysql_fetch_array($result)); //end while loop
$row = mysql_fetch_array($result);
1) remove this line of code from ur scripts
2) only use while loop code instead.

PHP Counting MYSQL rows issue?

I am trying to make it to if I have at least 1 team created for one person, it will display the team. If they don't have a team, it will say no teams. It works if the person has at least 1 team, but nothing shows if the person is not on a team. How do I fix this?
<?php
$sql = mysql_query("SELECT * FROM teams WHERE players LIKE '%$sessiongamt%'") or die("Could not allocate information!");
$num = 0;
while($row = mysql_fetch_assoc($sql)){
$num = ++$num;
$amount1 = mysql_num_rows($sql);
$name = $row["name"];
$teamrank = $row["rank"];
$teamlink = $row["link"];
$players = $row["players"];
$teamid = $row['id'];
if($amount1 < 1){
$teams = "No Teams";
echo "$amount";
}else{
$teams = "$name";
echo "<a href='$teamlink?id=$teamid'>$teams</a>";
}
}print "$amount1";
?>
Look at the third line of this code - put it in your code on the same place instead of your line.
if($amount1 < 1){
$teams = "No Teams";
echo $teams;
}else{
$teams = "$name";
echo "<a href='$teamlink?id=$teamid'>$teams</a>";
}
Put the counting outside the while... loop
$sql = mysql_query("SELECT * FROM teams WHERE players LIKE '%$sessiongamt%'") or die("Could not allocate information!");
$amount1 = mysql_num_rows($sql); //<---- this should fix it
$num = 0;
while($row = mysql_fetch_assoc($sql)){
$num = ++$num;
$name = $row["name"];
$teamrank = $row["rank"];
$teamlink = $row["link"];
$players = $row["players"];
$teamid = $row['id'];
$teams = "$name";
echo "<a href='$teamlink?id=$teamid'>$teams</a>";
}
if($amount1 < 1){
$teams = "No Teams";
}
print "$amount1";
?>

Listing the tables content in the database

written a script to print the contents of all the tables present in the database...
but unfortunately i am going wrong somewhere. could anyone please tell me where is the mistake?
$sql = "SHOW TABLES FROM $dbName";
$result = mysql_query($sql);
if (!$result) {
echo "DB Error, could not list tables\n";
echo 'MySQL Error: ' . mysql_error();
exit;
}
for ($i=1; $i<=5; $i++)
{
while ($row = mysql_fetch_row($result))
{
echo "Table: {$row[0]}\n";
$sql_1 = "SELECT * FROM {$row[0]}";
$result_1 = mysql_query($sql_1);
$row_1 = mysql_fetch_row($result);
echo "$row_1";
}
}
mysql_free_result($result);
?>
thanks..
first of all, u don't need
for ($i=1; $i<=5; $i++)
Also, $row_1 is an array, so to print it use something like
print_r($row_1);
And there might be not the only row in a table, so use construction like
while ($row = mysql_fetch_row($result))
{
echo "Table: {$row[0]}\n";
$sql_1 = "SELECT * FROM {$row[0]}";
$result_1 = mysql_query($sql_1);
while ($row_1 = mysql_fetch_row($result_1/*There was also an error*/))
print_r($row_1); /*Print_r or other function to print an array*/
}

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