This is my script but i need limit in mysql result i used before LIMIT in mysql query but it's not worked.
function display_menus($parent_id = 0){
$i=0;
$sql=mysql_query("SELECT * FROM customer_dog_details_db WHERE dog_id_parent = '$parent_id' && username='root' && dog_id_filter='$_GET[dog_id]'");
if(mysql_num_rows($sql) > 0){
echo "<ul class='tree'>";
while ($row = mysql_fetch_array($sql)){
echo "<li><a href=''><span style='font-size:14px;'>". $row['dog_name'] ."</span>";
echo "<br> ". $row['dog_id'] ." <br>";
echo "". $row['sex'] ."<br>";
echo "".$row['whelped_date']."<br>";
echo "</a>";
display_menus($row['dog_id']);
echo "</li>";
}
echo "</ul>";
}
}
display_menus();
use:
$sql=mysql_query("SELECT * FROM customer_dog_details_db WHERE dog_id_parent = '$parent_id' AND username='root' AND dog_id_filter='$_GET[dog_id]' limit 0,1");
instead of:
$sql=mysql_query("SELECT * FROM customer_dog_details_db WHERE dog_id_parent = '$parent_id' && username='root' && dog_id_filter='$_GET[dog_id]'");
Related
$sql = "SELECT * FROM today WHERE heading='$heading' and day='$day'";
$sql1 = "SELECT * FROM today WHERE day='$day'";
$result = $conn->query($sql);
$result1 = $conn->query($sql1);
if ($result->num_rows > 0) {
echo "<div id='post'><h1>".$row["heading"]."</h1>
<aside class='related-post'>".while($row = $result1->fetch_assoc())
{echo'<img src='".$row["image"]."'>;}
.</aside>}";
I have been using while loops for fetching data from table. My connection is working is perfect but I need another loop in the first that is not working. Isn't it the good way?
Update: I tried to finish echo and again started as follow but still an error
while($row = $result->fetch_assoc()) {
echo "<div id='post'><h1>"
.$row["heading"].
"</h1><div class='post-side'><img class='post-image' src='"
.$row["image"].
"'><div class='post-data'><p><strong>Age: </strong><span>$age</span></p><p><strong>Date of birth: </strong><span>"
.$row["day"].
"-"
.$row["month"].
"-"
.$row["year"].
"</span></p></div></div></div><div class='description'><p>"
.$row["description"].
"</p></div><div class='bottom-related'><aside class='related-post'>";
while($row = $result1->fetch_assoc())
{echo"<img src='"
.$row["image"].
"'>/";}.echo"</aside><aside class='ad2'>".$includead."</aside></div>";
}
echo "</div>";
} else {
echo "No table found";
}
$conn->close();
You're trying to concatenate to a string a WHILE loop; this is wrong.
You should echo your first part, end with it and then do your while loop, and echo the end afterwards:
Your quotes are a bit messed up as well
if ($result->num_rows > 0)
{
echo "<div id='post'><h1>".$row["heading"]."</h1>
<aside class='related-post'>";
while($row = $result1->fetch_assoc())
{
echo'<img src="'.$row["image"].'">';
}
echo '</aside>';
}
You can't concatene while with String, it's a syntaxic error
Also, you have a probleme when trying to echo a String, you can use this syntax:
echo "PHP"; // will evaluate PHP variables and whitespace inside a string
echo 'PHP'; // will evaluate nothing;
But you can not start flushing a string ' and finish by " or vice-versa.
Here the correct code :
<?php
$sql = "SELECT * FROM today WHERE heading='$heading' and day='$day'";
$sql1 = "SELECT * FROM today WHERE day='$day'";
$result = $conn->query($sql);
$result1 = $conn->query($sql1);
if ($result->num_rows > 0) {
echo "<div id='post'><h1>" . $row["heading"] . "</h1><aside class='related-post'>";
while($row = $result1->fetch_assoc()) {
echo'<img src="' . $row["image"] .'">';
}
echo "</aside>";
}
include "mysql.php";
$query= "SELECT ID,name,displayname,established,summary,searchlink,imagename,image FROM institutions ORDER BY rand() ";
$result=mysql_query($query,$db);
while($row=mysql_fetch_array($result))
{
echo "<div class='grid-item item1' style='background: ".ran_col().";'>";
$query1= "SELECT content FROM ".$row['name']." limit 1";
$result1=mysql_query($query1,$db);
while($row1=mysql_fetch_array($result1))
{
echo "<div class='content-short' data-id=".$row['name'].">";
$string = $row1['content'];
if (strlen($string) > 200)
{
$trimstring = substr($string, 0,200). '...';
}
else
{
$trimstring = substr($string,0). '...';
}
echo $trimstring;
echo "</div>";
}
echo <"/div">;
}
What code should do:
The code should grab data from institution table and after that $row['name'] should be used as table for next mysql query where from that $row['name'] table $row['content'] should be grabbed
but it is not working as expeted
You are using same variable everywhere. Change them to
include "mysql.php";
$query= "SELECT ID,name,displayname,established,summary,searchlink,imagename,image FROM institutions ORDER BY rand() ";
$result=mysql_query($query,$db);
while($row=mysql_fetch_array($result))
{
echo "<div class='grid-item item1' style='background: ".ran_col().";'>";
$query1= "SELECT content FROM ".$row['name']." limit 1";
$result1=mysql_query($query1,$db);
while($row1=mysql_fetch_array($result1))
{
echo "<div class='content-short' data-id=".$row['name'].">";
$string = $row1['content'];
if (strlen($string) > 200)
{
$trimstring = substr($string, 0,200). '...';
}
else
{
$trimstring = substr($string,0). '...';
}
echo $trimstring;
echo "</div>";
}
echo <"/div">;
}
EDIT:
Also your query is wrong
$query1= "SELECT content FROM ".$row['name']" limit 1";
to
$query1= "SELECT content FROM ".$row['name']." limit 1";
I'm pretty new at PHP/MySQL, so please be patient with me.
I am trying to get a list of members in a table to show up on a page. Right now it's showing the first member about 10 times and not displaying anyone else's name. I DID have it working, but I don't know what happened. I just want it to display everyone's name once. Here is my code:
<?php $select = mysql_query("SELECT * FROM `member_staff` WHERE `username`='$_SESSION[USR_LOGIN]' AND `status`='Active'");
$row = mysql_fetch_array($select);
$rows = mysql_num_rows($select);
$teaching = $row[teaching];
if ($rows==0){echo "Sorry, you don't appear to be a professor.";}
else { ?>
<?php }
$select2 = mysql_query("SELECT * FROM `classes_enrolled` WHERE `course`='" . $teaching . "' ORDER BY `student_name`") or die(mysql_error());
$count = mysql_num_rows($select2);
$row2 = mysql_fetch_array($select2);
$student=$row2[student_name];
if($count==NULL) {
echo "<table width=\"80%\">\n";
echo "<tr><td><center>Nobody has registered for your class yet!</center></td></tr>\n";
echo "</table>\n";
echo "<br /><br />\n\n";
}
else {
echo "<center><font size=\"3\"><b>YEAR 1, TERM 2</b></font></center>";
echo "<table width=\"80%\" class=\"table-stripes\">\n";
echo "<tr><td width=\"50%\"><b>STUDENT</b></td></tr>\n";
$select3 = mysql_query("SELECT * FROM `members` WHERE `username`='" . $student . "'") or die(mysql_error());
$row3 = mysql_fetch_array($select3);
while($row2 = mysql_fetch_array($select2)) {
$house=$row3[house];
echo "<tr><td><strong class=\"$house\">$student</strong></td></tr>";
}
echo "</table>"; }
?>
I miss look on your code, since it is mess, but disregard the mysqli and mysql thing, you want to show how many student in the teacher's classes.
<?php $select = mysql_query("SELECT * FROM `member_staff` WHERE `username`='$_SESSION[USR_LOGIN]' AND `status`='Active'");
$row = mysql_fetch_array($select);
$rows = mysql_num_rows($select);
$teaching = $row[teaching]; <--- This only get first row of the course, if you want multiple course under same username, you need to loop it.
if ($rows==0){echo "Sorry, you don't appear to be a professor.";}
else { ?>
<?php }
$select2 = mysql_query("SELECT * FROM `classes_enrolled` WHERE `course`='" . $teaching . "' ORDER BY `student_name`") or die(mysql_error());
$count = mysql_num_rows($select2);
$row2 = mysql_fetch_array($select2);
$student=$row2[student_name]; <----- This only get the first row of the student name, if you want multiple student under a course, you need to loop it.
if($count==NULL) {
echo "<table width=\"80%\">\n";
echo "<tr><td><center>Nobody has registered for your class yet!</center></td></tr>\n";
echo "</table>\n";
echo "<br /><br />\n\n";
}
else {
echo "<center><font size=\"3\"><b>YEAR 1, TERM 2</b></font></center>";
echo "<table width=\"80%\" class=\"table-stripes\">\n";
echo "<tr><td width=\"50%\"><b>STUDENT</b></td></tr>\n";
$select3 = mysql_query("SELECT * FROM `members` WHERE `username`='" . $student . "'") or die(mysql_error());
$row3 = mysql_fetch_array($select3);
while($row2 = mysql_fetch_array($select2)) {
$house=$row3[house]; <----This only show the first row of $house under same student, so you need to loop it too.
echo "<tr><td><strong class=\"$house\">$student</strong></td></tr>";
}
echo "</table>"; }
?>
So what you really want to do is
<?php
$select = mysql_query("SELECT * FROM `member_staff` WHERE `username`='$_SESSION[USR_LOGIN]' AND `status`='Active'");
$rows = mysql_num_rows($select);
if ($rows==0){echo "Sorry, you don't appear to be a professor.";}
else { ?>
<?php }
while( $row = mysqli_fetch_array( $select ) ) {
$teaching = $row[teaching];
$select2 = mysql_query("SELECT * FROM `classes_enrolled` WHERE `course`='" . $teaching . "' ORDER BY `student_name`") or die(mysql_error());
$count = mysql_num_rows($select2);
if($count==NULL) {
echo "<table width=\"80%\">\n";
echo "<tr><td><center>Nobody has registered for your class yet!</center></td></tr>\n";
echo "</table>\n";
echo "<br /><br />\n\n";
} else {
while( $row2 = mysql_fetch_array($select2) ) {
$student=$row2[student_name];
echo "<center><font size=\"3\"><b>YEAR 1, TERM 2</b></font></center>";
echo "<table width=\"80%\" class=\"table-stripes\">\n";
echo "<tr><td width=\"50%\"><b>STUDENT</b></td></tr>\n";
$select3 = mysql_query("SELECT * FROM `members` WHERE `username`='" . $student . "'") or die(mysql_error());
while($row3 = mysql_fetch_array($select3)) {
$house=$row3[house];
echo "<tr><td><strong class=\"$house\">$student</strong></td></tr>";
}
echo "</table>";
}
} // END ELSE
}
} // END ELSE
?>
I am trying to display a list of links based upon mysql output. The code works except that it drops the first record. I know it is because of the duplicate [ #row3 = mysql_fetch_array($result3) ] entries but if i remove one the code fails. Can anyone suggest a fix?
Thanks
<?php
$sql3 = "SELECT `Record_ID`, `Name` FROM `rides` WHERE `Rating` = 3";
$result3=mysql_query($sql3)or die(mysql_error());
//var_dump ($result3);
$num = mysql_num_rows($result3);
while ($row3 = mysql_fetch_array($result3))
{
echo "<table>";
for ($i = 0; $i < $num; $i++){
$row3 = mysql_fetch_array($result3);
//var_dump($row3);
$ridesid = $row3[0];
$rides = $row3[1];
echo "<tr>";
echo "<a href='attraction_page.php?rideID=". urlencode($ridesid) ."'>$rides</a>";
echo "<br />";
echo "</tr>";
}
echo '</table>';
}
?>
You have fetch same thing twice!
Try this:
<?php
$sql3 = "SELECT `Record_ID`, `Name` FROM `rides` WHERE `Rating` = 3";
$result3=mysql_query($sql3)or die(mysql_error());
//var_dump ($result3);
$num = mysql_num_rows($result3);
echo "<table>";
while ($row3 = mysql_fetch_array($result3))
{
$ridesid = $row3[0];
$rides = $row3[1];
echo "<tr>";
echo "<a href='attraction_page.php?rideID=". urlencode($ridesid) ."'>$rides</a>";
echo "<br />";
echo "</tr>";
}
echo '</table>';
?>
Call mysql_fetch_array() only once...
How would I do let the user know if there wasnt any rows found?
And if it finds (for example) 26 rows I want it to print that it found 26 hits (Your search for "hey" gave 26 results)
$name = mysql_real_escape_string($_POST['player_name']);
$q = mysql_query("SELECT * FROM players WHERE name LIKE '%$name%'");
while ($row = mysql_fetch_array($q)) {
$name = $row['name'];
echo "<ul>\n";
echo "<li>" . "" .$name . " </li>\n";
echo "</ul>";
}
Use mysql_num_rows:
$rowCount = mysql_num_rows($q);
if ($rowCount === 0)
{
echo 'Found nothing!';
}
else
{
echo "Found $rowCount rows!";
while ($row = mysql_fetch_array($q))
{
$name = $row['name'];
echo "<ul>\n";
echo "<li>" . "" .$name . " </li>\n";
echo "</ul>";
}
}
Check out mysql_num_rows
mysql_num_rows
$numberOfResults = mysql_num_rows($q);