Hiii, I am getting multiple value form database when I am using foreach, plz help
function display($host,$user,$pass,$database)
{
$db = mysql_connect($host, $user, $pass);
mysql_select_db ($database);
$query = "SELECT * FROM `sysdes_moduleinfo`";
$result = mysql_query($query) OR die(mysql_error());
$i=0;
while($row = mysql_fetch_array($result))
{
/*$max = count($row);
while($i<6) {
echo $row[$i]." ";
$i++;
}*/
foreach ($row as $value)
{
//echo $value . " ";
echo htmlspecialchars($value);
}
echo "<br/>";
}
This what I get with this code.
.
This what I have in the database.
The default result type for mysql_fetch_array is to return data in both a number and associative array.
This is why data is duplicated. Try using mysql_fetch_row instead.
However it should be noted that both mysql_fetch_array & mysql_fetch_row are both deprecated.
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
foreach ($row as $value)
{
//echo $value . " ";
echo htmlspecialchars($value);
}
echo "<br/>";
}
mysql_fetch_array returns both a numeric and associative array.
Related
This shows all the elements from an array in a string but without a separator. I used ',' as a separator in this code. And it does not work. How can i separate them with separator.
$con = mysql_connect("localhost","root","");
if(!$con)
{
die("Database Connection failed.".mysql_error());
}
$db = mysql_select_db("test1",$con);
if(!$db)
{
die("Database selection failed.".mysql_error());
}
$query = "select * from menu limit 10";
$result = mysql_query($query,$con);
while($row = mysql_fetch_array($result))
{
$id = $row['id'];
//$menu_name = $row['m_name'];
//$menu_image = $row['m_image'];
$menu_name = array($row['m_name']);
$menu_image = array($row['m_image']);
echo implode(',',$menu_name);
//echo "<img src='images/$menu_image' style='height:200px;width:200px;'>";
}
if(!$result)
{
echo mysql_error;
}
?>
you are storing array into variable . so if you will use imploade on an variable it will not show the result so best way is to store your result into an array and then use implode .
Don't use depricated mysql_* function use mysqli_* or pdo
instead
$menu_image=array(); // start array to store
$menu_name=array();
while($row = mysql_fetch_array($result))
{
$id = $row['id'];
$menu_name[] = $row['m_name']; // store menu name
$menu_image[] = $row['m_image']; // store menu image
}
echo implode(',',$menu_name); // finaly use implode
You can try this:
$menu_name[];
$menu_image[];
while($row = mysql_fetch_array($result)) {
$id = $row['id'];
$menu_name[] = $row['m_name']; // I change your $menu_name into an array so that when you fetch $row['m_name'] it returns an array
$menu_image[] = $row['m_image']; //same thing
}
echo implode(',',$menu_name);
//echo "<img src='images/$menu_image' style='height:200px;width:200px;'>";
try this in place of your while loop
$menu_name = array();
$menu_image = array();
while($row = mysql_fetch_array($result))
{
$id = $row['id'];
$menu_name[] = $row['m_name'];
$menu_image[] = $row['m_image'];
}
echo implode(',',$menu_name);
The code is supposed to get a row at each loop and build td elements for each data piece. However, it only gets some rows while missing others even though all rows were selected:
$query = "select * from category";
$ret = mysql_query($query, $conn);
while ($rows = mysql_fetch_assoc($ret)) {
echo "<tr>";
foreach ($rows as $val) {
echo "<td>{$val}</td>";
}
}
mysql_fetch_assoc returns an associative array. To display each returned row, do something like this (taken directly from the php.net page for mysql_fetch_assoc
while ($row = mysql_fetch_assoc($result)) {
echo $row["userid"];
echo $row["fullname"];
echo $row["userstatus"];
}
Try removing the curly braces like this:
echo "<td>$val</td>";
Beside you didn't close the <tr> tag. So you should do this:
$query = "select * from category";
$ret = mysql_query($query, $conn);
while ($rows = mysql_fetch_assoc($ret)) {
echo "<tr>";
foreach ($rows as $val) {
echo "<td>$val</td>";
}
echo"</tr>";
}
Now like Jack Albright said you should consider various columns of your table to display specific data. Upon that the while() is already a look, I think you don't need to add any foreach() inside. SO your final code should look like this:
$query = "select * from category";
$ret = mysql_query($query, $conn);
while ($rows = mysql_fetch_assoc($ret)) {
echo "<tr>";
echo "<td>$rows['columnName']</td>";
echo "</tr>";
}
I have this code...
$results_query = "SELECT * FROM profile WHERE string LIKE '%" . $search_deliminator . "%'";
$results_result = $database->query($results_query);
$result = mysql_fetch_array($results_result);
What I'm trying to do is pull all the rows in which the string column contains my search deliminator. Further, I would like to get the values from each column of the rows.
How would I pull the multidimensional array I need containing each row, and each value of each column within each row?
EDIT:
I'm looking to do something to this effect...
while ($row = mysql_fetch_assoc($results_result)) {
$result[] = $row;
}
Then echo each column like this...
foreach ($result as $row) {
echo $row["0"];
}
To get a flattened array of the result set, try this:
$result = array();
while ($row = mysql_fetch_assoc($results_result)) {
$result[$row['id']] = $row;
}
echo json_encode($result);
Also, you should look into either MySQLi or PDO to replace the (now deprecated) MySQL extension.
If you use PDO, you can use the fetchAll() method:
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
With mysql or mysqli, you use a loop:
$rows = array();
while ($row = $stmt->fetch_assoc()) {
$rows[] = $row;
}
try this
$i=0;
while ($row = mysql_fetch_assoc($results_result)) {
$result[$i]["column1"] = $row["column1"];
$result[$i]["column2"] = $row["column2"];
$i++;
}
To display the output use:
foreach ($result as $row) {
echo $row["column1"];
echo $row["column2"];
}
ive got a script in php which fetched data from server, but i want to encode it in json format.
<?php
// attempt a connection
$dbh = pg_connect("host=10.22.35.11 dbname=iwmp_dev2 user=postgres ");
if (!$dbh) {
die("Error in connection: " . pg_last_error());
}
// execute query
//$sql = $_POST['pLat'];
$sql = "SELECT officer_name FROM iwmp_officer";
$result = pg_query($dbh, $sql);
if (!$result) {
die("Error in SQL query: " . pg_last_error());
}
$array = array();
while ($row = pg_fetch_assoc($result)) {
$i++;
$comm = implode(",",$row);
echo json_encode($comm);
}
// free memory
pg_free_result($result);
// close connection
pg_close($dbh);
?>
but my ouptput is coming in format
"V. M. ARORA""Dr. C. P. REDDY""ARTI CHOWDHARY""JAGDISH SINGH"
also when using implode func on *pgsql_fetch_assoc*, no ","(coma's) are coming.
please help
I think you're doing it in wrong way. Try to do like below.
while ($row = pg_fetch_assoc($result)) {
$rows[] = $row;
}
echo json_encode($rows);
Try this;
while ($row = pg_fetch_assoc($result)) {
echo json_encode($row);
}
You don't need to implode the $row. Try replacing your while loop with below code
while ($row = pg_fetch_assoc($result)) {
echo json_encode($row);
}
i want to echo out everything from a particular query. If echo $res I only get one of the strings. If I change the 2nd mysql_result argument I can get the 2nd, 2rd etc but what I want is all of them, echoed out one after the other. How can I turn a mysql result into something I can use?
I tried:
$query="SELECT * FROM MY_TABLE";
$results = mysql_query($query);
$res = mysql_result($results, 0);
while ($res->fetchInto($row)) {
echo "<form id=\"$row[0]\" name=\"$row[0]\" method=post action=\"\"><td style=\"border-bottom:1px solid black\">$row[0]</td><td style=\"border-bottom:1px solid black\"><input type=hidden name=\"remove\" value=\"$row[0]\"><input type=submit value=Remove></td><tr></form>\n";
}
$sql = "SELECT * FROM MY_TABLE";
$result = mysqli_query($conn, $sql); // First parameter is just return of "mysqli_connect()" function
echo "<br>";
echo "<table border='1'>";
while ($row = mysqli_fetch_assoc($result)) { // Important line !!! Check summary get row on array ..
echo "<tr>";
foreach ($row as $field => $value) { // I you want you can right this line like this: foreach($row as $value) {
echo "<td>" . $value . "</td>"; // I just did not use "htmlspecialchars()" function.
}
echo "</tr>";
}
echo "</table>";
Expanding on the accepted answer:
function mysql_query_or_die($query) {
$result = mysql_query($query);
if ($result)
return $result;
else {
$err = mysql_error();
die("<br>{$query}<br>*** {$err} ***<br>");
}
}
...
$query = "SELECT * FROM my_table";
$result = mysql_query_or_die($query);
echo("<table>");
$first_row = true;
while ($row = mysql_fetch_assoc($result)) {
if ($first_row) {
$first_row = false;
// Output header row from keys.
echo '<tr>';
foreach($row as $key => $field) {
echo '<th>' . htmlspecialchars($key) . '</th>';
}
echo '</tr>';
}
echo '<tr>';
foreach($row as $key => $field) {
echo '<td>' . htmlspecialchars($field) . '</td>';
}
echo '</tr>';
}
echo("</table>");
Benefits:
Using mysql_fetch_assoc (instead of mysql_fetch_array with no 2nd parameter to specify type), we avoid getting each field twice, once for a numeric index (0, 1, 2, ..), and a second time for the associative key.
Shows field names as a header row of table.
Shows how to get both column name ($key) and value ($field) for each field, as iterate over the fields of a row.
Wrapped in <table> so displays properly.
(OPTIONAL) dies with display of query string and mysql_error, if query fails.
Example Output:
Id Name
777 Aardvark
50 Lion
9999 Zebra
$result= mysql_query("SELECT * FROM MY_TABLE");
while($row = mysql_fetch_array($result)){
echo $row['whatEverColumnName'];
}
$sql = "SELECT * FROM YOUR_TABLE_NAME";
$result = mysqli_query($conn, $sql); // First parameter is just return of "mysqli_connect()" function
echo "<br>";
echo "<table border='1'>";
while ($row = mysqli_fetch_assoc($result)) { // Important line !!!
echo "<tr>";
foreach ($row as $field => $value) { // If you want you can right this line like this: foreach($row as $value) {
echo "<td>" . $value . "</td>";
}
echo "</tr>";
}
echo "</table>";
In PHP 7.x You should use mysqli functions and most important one in while loop condition use "mysqli_fetch_assoc()" function not "mysqli_fetch_array()" one. If you would use "mysqli_fetch_array()", you will see your results are duplicated. Just try these two and see the difference.
Nested loop to display all rows & columns of resulting table:
$rows = mysql_num_rows($result);
$cols = mysql_num_fields($result);
for( $i = 0; $i<$rows; $i++ ) {
for( $j = 0; $j<$cols; $j++ ) {
echo mysql_result($result, $i, $j)."<br>";
}
}
Can be made more complex with data decryption/decoding, error checking & html formatting before display.
Tested in MS Edge & G Chrome, PHP 5.6
All of the snippets on this page can be dramatically reduced in size.
The mysqli result set object can be immediately fed to a foreach() (because it is "iterable") which eliminates the need to maked iterated _fetch() calls.
Imploding each row will allow your code to correctly print all columnar data in the result set without modifying the code.
$sql = "SELECT * FROM MY_TABLE";
echo '<table>';
foreach (mysqli_query($conn, $sql) as $row) {
echo '<tr><td>' . implode('</td><td>', $row) . '</td></tr>';
}
echo '</table>';
If you want to encode html entities, you can map each row:
implode('</td><td>' . array_map('htmlspecialchars', $row))
If you don't want to use implode, you can simply access all row data using associative array syntax. ($row['id'])