I don't know why doesn't display the avatar 100x100. Something is wrong and I don't know how to fix it.
$result = mysqli_query($con,"SELECT * FROM `users` WHERE `verified` = 1 and lastcheck=0 order by `authoredPostCount` DESC");
echo "<table border='1'> <tr>
<th>Picture</th>
<th>Verified</th>
<th>Videos</th></tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row "<img src="['avatar']."></td>";
echo "<td>" . $row['userId'] . "</td>";
echo "<td>" . $row['authoredPostCount'] . "</td>";
echo "</tr>";
}
echo "</table>";
if (!$cuserId) {
printf("Error: %s\n", mysqli_error($con));
exit();
}
mysqli_close($con);
?>
Replace
echo "<td>" . $row "<img src="['avatar']."></td>";
With
echo "<td><img src=". $row['avatar']."></td>";
if you are saving image file name in database then use it like that
echo "<td><img src='path/to/folder/". $row['avatar'].".jpeg'></td>";
Related
Two questions in one:
First question, mainly about presentation:
I'm echo-ing the following code which should create a table. The table should have a single column, but it's being rendered with the elements above the image as a single line. can anyone see why?
<?php
$sql = "SELECT * FROM catdata WHERE featured='yes' LIMIT 2";
if($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result) > 0){
echo "<table>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['manufacturer'] . "</td>";
echo "</tr";
echo "<tr>";
echo "<td>" . $row['title'] . "</td>";
echo "</tr";
echo "<tr>";
echo "<td><img src=\"6.diesel.png\"></td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row['size'] . "l</td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row['mileage'] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row['fitsmodel'] . "</td>";
echo "</tr>";
echo "<tr class=\"tablePriceBlock\">";
echo "<td>£" . $row['pricefitted'] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td>£" . $row['pricedel'] . "</td>";
echo "</tr>";
}
echo "</table>";
// Free result set
mysqli_free_result($result);
} else {
echo "No records matching your query were found.";
}
} else{
echo "ERROR: Unable to execute $sql. " . mysqli_error($link);
}
?>
Question 2:
Can I show the second result in a second column, or as a separate table? Also, is it possible to access the $result elements like say, $[manufacturer][1]?
Always look at the emitted source your code generates by either "View Source" or using a tool like curl. You'll find this mistake:
echo "</tr";
You're missing a >.
Many people who write HTML have browser plugins that can link through to an HTML validator to ensure they've got the correct syntax. You may want to find and install one of these.
To generate the second result is a separate table follow the following code.
echo "<table>";
$count = 1;
while($row = mysqli_fetch_array($result)){
if($count == 2){
echo "<table>";
echo "<tr>";
echo "<td>put your code here</td>";
echo "</tr>"
echo </table>
}else{
echo "<tr>";
echo "<td>" . $row['manufacturer'] . "</td>";
echo "</tr";
echo "<tr>";
echo "<td>" . $row['title'] . "</td>";
echo "</tr";
echo "<tr>";
echo "<td><img src=\"6.diesel.png\"></td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row['size'] . "l</td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row['mileage'] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row['fitsmodel'] . "</td>";
echo "</tr>";
echo "<tr class=\"tablePriceBlock\">";
echo "<td>£" . $row['pricefitted'] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td>£" . $row['pricedel'] . "</td>";
echo "</tr>";
}
$count ++;
}
echo "</table>";
As you are limiting your query to only two record so this method is not bad for that and hope this will help.
This is my code (horrible one):
<?php
include 'connect/con.php';
$result = mysqli_query($con,"SELECT id, vidTitle FROM newsvid");
$result1 = mysqli_query($con,"SELECT imgCover, vidSD FROM newsvid");
$result2 = mysqli_query($con,"SELECT published FROM newsvid");
echo "<table width=\"600\" border=\"1\"><tbody>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo '<td width=\"10%\">'.$row['id'].'</td>';
echo "<td width=\"90%\">" . $row['vidTitle'] . "</td>";
echo "</tr>";
}
echo "</tbody></table>";
echo "<table width=\"600\" border=\"1\"><tbody>";
while($row = mysqli_fetch_array($result1)) {
echo "<tr>";
echo "<td width=\"40%\">" . $row['imgCover'] . "</td>";
echo "<td width=\"60%\">" . $row['vidSD'] . "</td>";
echo "</tr>";
}
echo "</tbody></table>";
echo "<table width=\"600\" border=\"1\"><tbody>";
while($row = mysqli_fetch_array($result2)) {
echo "<tr>";
echo "<td >" . $row['published'] . "</td>";
echo "</tr>";
}
echo "</tbody></table>";
mysqli_close($con);
?>
</body>
</html>
The question is how to show data from database in this layout:
--------------------------------
-id----------vidTitle-----------
--------------------------------
-imgCover------vidSD------------
--------------------------------
----------published-------------
So every time I will add more data , another block like I showed before will add up under existing one.
........................................................................................
There's no need to write 3 queries. You could do that with only one select, and then put all the echos inside a while. That way you're writing, it would run all the ids and titles first, then it would put a table after the table, with cover and vidSD.
Try to make a single query:
SELECT id, vidTitle, imgCover, vidSD, published FROM newsvid
That way you will have, on each row returned from database, all the information about the same row.
Now, running a while is the same as you're doing, just adapting some HTML:
echo "<table width='600' border='1'><tbody>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo '<td width=\"10%\">'.$row['id'].'</td>';
echo "<td width=\"90%\">" . $row['vidTitle'] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td width=\"40%\">" . $row['imgCover'] . "</td>";
echo "<td width=\"60%\">" . $row['vidSD'] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td colspan='2'>" . $row['published'] . "</td>";
echo "</tr>";
}
echo "</tbody></table>";
You may want to order it too. Adding ORDER BY id DESC, would do that.
I'm trying to assign anID to the table I'm generating with PHP but it keeps returning an error. Here is the full code that works fine so far. All I want is to add an 'id' to the table so I can apply css styles to it in the relevant sheet
<?php
$con=mysqli_connect("localhost","<un>","<pw>","monitor");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM presnationalresults ORDER BY Percentage DESC");
echo "<table border='1'>
<tr>
<th>President</th>
<th>Party</th>
<th>Votes</th>
<th>Percentage</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['PresidentName'] . "</td>";
echo "<td>" . $row['PartyCode'] . "</td>";
echo "<td>" . $row['Votes'] . "</td>";
echo "<td>" . $row['Percentage'] . "</td>";
}
echo "</table>";
mysqli_close($con);
Any help? maybe I'm going about it the wrong way?
Please try below block of code . I have added table id and also close </tr> inside while loop
<?php
$con = mysqli_connect("localhost", "<un>", "<pw>", "monitor");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con, "SELECT * FROM presnationalresults ORDER BY Percentage DESC");
echo "<table border='1' id='table-id'>
<tr>
<th>President</th>
<th>Party</th>
<th>Votes</th>
<th>Percentage</th>
</tr>";
while ($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['PresidentName'] . "</td>";
echo "<td>" . $row['PartyCode'] . "</td>";
echo "<td>" . $row['Votes'] . "</td>";
echo "<td>" . $row['Percentage'] . "</td>";
echo "</tr>"; // you forget close tr
}
echo "</table>";
mysqli_close($con);
I have MAMP (local hosted SQL,WEB etc server) the database name is :NKTDEBITS the table name is :Insurance and the column on the table is STATECOV. I know I'm close with this but still get a black in the field that should generate the total, anyone got a idea?
<?php
$con=mysqli_connect("localhost","root","root","KNTDebits");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM Insurance");
$result2 = mysqli_query($con,"SELECT * FROM Office");
$result3 = mysqli_query($con,"SELECT * FROM RichmondLocation");
$result4 = mysqli_query($con,"SELECT * FROM DanvilleLocation");
$result5 = mysql_query('SELECT SUM(STATECOV) AS STATECOV_sum FROM Insurance');
echo "<table border='1'>
<tr>
<th>Truck Number</th>
<th>VIN</th>
<th>Make</th>
<th>Model</th>
<th>State Coverage</th>
<th>Comprehinsive Coverage</th>
<th>Property Damage/th>
<th>Personal Injury</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['TNUM'] . "</td>";
echo "<td>" . $row['VIN'] . "</td>";
echo "<td>" . $row['MAKE'] . "</td>";
echo "<td>" . $row['MODEL'] . "</td>";
echo "<td>" . $row['STATECOV'] . "</td>";
echo "<td>" . $row['COMPRE'] . "</td>";
echo "<td>" . $row['PROPDMG'] . "</td>";
echo "<td>" . $row['PRSINJ'] . "</td>";
echo "</tr>";
}
echo "</table>";
//Table 2 Start
str_repeat(' ', 5); // adds 5 spaces
echo "<table border='5'>
<tr>
<th>Richmond</th>
<th>Date</th>
<th>Payment</th>
<th>Payer</th>
</tr>";
while($row3 = mysqli_fetch_array($result3))
{
echo "<tr>";
echo "<td>" . $row3[''] . "</td>";
echo "<td>" . $row3['DATE'] . "</td>";
echo "<td>" . $row3['PAYMENT'] . "</td>";
echo "<td>" . $row3['PAYER'] . "</td>";
echo "</tr>";
}
echo "</table>";
//Table 4 Start
str_repeat(' ', 5); // adds 5 spaces
echo "<table border='5'>
<tr>
<th>Danville</th>
<th>Date</th>
<th>Payment</th>
<th>Payer</th>
</tr>";
while($row4 = mysqli_fetch_array($result4))
{
echo "<tr>";
echo "<td>" . $row4[''] . "</td>";
echo "<td>" . $row4['DATE'] . "</td>";
echo "<td>" . $row4['PAYMENT'] . "</td>";
echo "<td>" . $sum . "</td>";
echo "</tr>";
}
echo "</table>";
//Table 5 Start
echo "<table border='5'>
<tr>
<th>Total</th>
</tr>";
$result = mysql_query('SELECT SUM(STATECOV) AS value_sum FROM Insurance');
$row = mysql_fetch_assoc($result);
$sum = $row['value_sum'];
while($row = mysql_fetch_assoc($result));
{
echo "<tr>";
echo "<td>" . $sum . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
You need a GROUP BY to go along with your SUM. I don;t know enough about you table to let you know what column you should use for this.
You should be handling potential error cases whenn you query the database, as you will quickly see when you are getting database/query errors.
You should also look to use mysqli or PDO instead of the deprecated mysql_* functions.
I want to create a compare sort of page. I currently have two tables with the information, but want it to be on table or join both together. The first row would show Name of brand, second would show cost, etc... I am grabbing the values of selected checkboxs and based on what they selected it get compared.
$testName = $_POST['testCompare'];
$rpl = str_replace("_", " ", $testName);
$firstOne = "$rpl[0]";
$secondOne = "$rpl[1]";
echo "$firstOne";
echo "<br/>";
echo "$secondOne";
$query = "SELECT * FROM test_table WHERE test_name = '$firstOne'";
$query2 = "SELECT * FROM test_table WHERE test_name = '$secondOne'";
$result = mysql_query($query) or die ("Error in query: $query. " . mysql_error());
$result2 = mysql_query($query2) or die ("Error in query: $query2. " . mysql_error());
if (mysql_num_rows($result) > 0 && mysql_num_rows($result2) > 0) {
//if (mysql_num_rows($result) > 0) {
while($row = mysql_fetch_row($result)) {
if ($firstOne == $row[1]) {
{
echo "<table border=1>";
echo "<tr>";
echo "<td>" . $row[0] . "</td>";
echo "<tr>";
echo "<td>" . $row[1] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row[2] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row[3] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row[4] . "</td>";
echo "</tr>";
}
}
}echo "</table>";
while($row2 = mysql_fetch_row($result2)) {
if ($secondOne == $row2[1]) {
{ echo "<table border=1>";
echo "<tr>";
echo "<td>" . $row2[0] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row2[1] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row2[2] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td>" . $row2[3] . "</td>";
echo "</tr>";
echo "<td>" . $row2[4] . "</td>";
echo "</tr>";
}
}
}
echo "</table>";
}
else {
echo "No Results";
}[/CODE]
Thanks
Remove the </table> after the first loop.
Remove <table border=1> from both loops.
Add <table border=1 before the first loop.
Currently, you're defining a new <table border=1> each time you enter the loop. This will result in this HTML code:
<table ..>
<tr>...
<table ..>
..
<table>
..
Et cetera
</table>
<table ..>
Et cetera
</table>