Any ideas why this wouldnt work!?
Just nothing displays..
$result1 = mysql_query("select * from `suppliers` where supp_short_code='WED'") or die(mysql_error());
while($row1 = mysql_fetch_array($result1))
{
echo "<p>" . $row1['supp_name'] . "</p>";
}
Resolved! turns out when i imported from Excel to my mysql in brought in spaces after and before the "supp_short_code" so the query wasnt working!
Is 'supp_name' a column of your 'suppliers' table?
Does this query return any result under phpmyadmin eg?
Are the <p> tags displayed (no rules as display:none;)?
Where is your connection? If there isn't one, mysql_query() tries to make one by calling mysql_connect() with no args. It also fails if your connection doesn't have permissions for the table you're trying to read. If it's not dying but only returning no result, then check mysql_num_rows()
correct quotes of this line ::
$result1 = mysql_query("select * from suppliers where supp_short_code='WED'") or die(mysql_error());
Related
My query is not working when I use the variable in the WHERE clause. I have tried everything. I echo the variable $res, it shows me the perfect value, when I use the variable in the query the query is not fetching anything thus mysqli_num_rows is giving me the zero value, but when I give the value that the variable contains statically the query executes perfectly. I have used the same kind of code many times and it worked perfectly, but now in this part of module it is not working.
Code:
$res = $_GET['res']; // I have tried both post and get
echo $res; //here it echos the value = mahanta
$query = "SELECT * FROM `seller` WHERE `restaurant` = '$res'"; // Here it contains the problem I have tried everything. Note: restaurant name is same as it is in the database $res contains a value and also when I give the value of $res i.e. mahanta in the query it is then working.
$z = mysqli_query($conn, $query);
$row2 = mysqli_fetch_array($z);
echo var_dump($row2); // It is giving me null
$num = mysqli_num_rows($z); // Gives zero
if ($num > 0) {
while ($row2 = mysqli_fetch_array($z)) {
$no = $row2['orders'];
$id = $res . $no;
}
}
else {
echo "none selected";
}
As discussed in the comment. By printing the query var_dump($query), you will get the exact syntax that you are sending to your database to query.
Debugging Tip: You can also test by pasting the var_dump($query) value in your database and you will see the results if your query is okay.
So update your query syntax and print the query will help you.
$query = "SELECT * FROM `seller` WHERE `restaurant` = '$res'";
var_dump($query);
Hope this will help you and for newbies in future, how to test your queries.
Suggestion: Also see how to write a mysql query syntax for better understanding php variables inside mysql query
The problem is the way you're using $res in your query. Use .$res instead. In PHP (native or framework), injecting variables into queries need a proper syntax.
I'm trying to do a simple query:
SELECT
job_title,
job_description,
job_tasks,
technology_skills,
job_activities
FROM
careeryou_db3
WHERE
Job_Interests = " CIE"
It works perfectly in mysql giving me this result:
http://imgur.com/a/E9yv3
However, when I use this in my PHP it returns an empty data set. I did some troubleshooting by trying to query the columns one by one and it works(returning the correct results respectively.
SELECT
job_title, job_description, job_tasks
FROM
careeryou_db3
WHERE
Job_Interests = " CIE"
Apparently, when I try to query anything from the columns of [technology_skills] OR [job_activities], the php result would return as BLANK. Does anyone have any idea why? this is strange and im kinda new to this
This is my php code:
$query = 'SELECT job_title,job_description,job_tasks FROM careeryou_db3 WHERE Job_Interests = " CIE" ';
$resultset = mysql_query($query, $connection);
$records = array();
while($r = mysql_fetch_assoc($resultset)){
$records[] = $r;
}
echo json_encode($records);
I can't see any error on your sql syntax. Can you check the mysql logs for better error information?
See here for details
https://dev.mysql.com/doc/refman/5.7/en/error-log.html
[mysqld]
log_error=/var/log/mysql/mysql_error.log
general_log_file=/var/log/mysql/mysql.log
It would help how you call the mysql, place the query build code as example.
It might be worth adding these lines for extra debugging to ensure your connection and query is successful.
$result = mysql_query($sql);
if (!$result) {
echo "Could not successfully run query ($sql) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($result) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
while ($row = mysql_fetch_assoc($result)) {
....
It is because of the data in those columns that are conflicting with the PHP. Apparently this "—" cannot get parse into PHP which explains why the query works only in phpMyAdmin and not my PHP script.
I do a query but does not show anything on the screen and when I opened the page where I do the query is slow and does not show anything
$query="SELECT * FROM Dettagli_macchina WHERE macchine_id='$macchine' and Email='$_SESSION[login_user]'";
$result = mysqli_query($conne,$query);
while($row=mysqli_num_rows($result)){
echo $row['COMPONENTE'];
}
mysqli_num_rows() will only use for getting no of rows not for row data.
You need yo use mysqli_fetch_*()
while($row=mysqli_fetch_array($result)){
echo $row['COMPONENTE'];
}
Why this query slow? Because you are using infinite loop here, always TRUE.
while($row=mysqli_num_rows($result))
One more story, I hope you are using session_start() in your file, but suppose that if $_SESSION not found or not start than your query will failed.
In last, this is just a suggestion regarding Naming Convention, you are using column name in small letter, capital small, full capital, this is not related to answer but you must need to learn about this art.
this will help you to understand Naming Convention: Database, Table and Column Naming Conventions?
This reference will help you to understand how mysqli_fetch_array() works: http://php.net/manual/en/mysqli-result.fetch-array.php
Please try with this one for return value and if you need number of raw then the statements will be different.
$query="SELECT * FROM Dettagli_macchina WHERE macchine_id='$macchine' and Email='$_SESSION[login_user]'";
$result = mysqli_query($conne,$query);
while($row=mysqli_fetch_array($result)){
echo $row['COMPONENTE'];
}
As others have pointed out, you have mixed up the mysqli_num_rows function, here I am using it to print the number of results found, then loop through the results after converting the mysqli result object to an array named $row
echo 'Found '. mysqli_num_rows($result) .' results';
while ($row = mysqli_fetch_array($result)) {
echo $row['COMPONENTE'];
}
You are fetching record but using mysqli_num_rows() which return total number of rows/records replace it with mysqli_fetch_array() here is the working example.
$query="SELECT * FROM Dettagli_macchina WHERE macchine_id='$macchine' and Email='$_SESSION[login_user]'";
$result = mysqli_query($conne,$query);
while($row = mysqli_fetch_array($result)) { //replace mysqli_num_rows with mysqli_fetch_array
echo $row['COMPONENTE'];
}
I have a problem. I have an array of values from database, when I try to pass it to a string with commas, it works fine on my localhost, but when I upload it to my online server, the string doesn't show any values. For example: select from table where in (,,) only shows the commas and in my xampp server it works excellent. Any ideas what this can be?
Here's the code:
<?php
$sql = "select id from users where gid = 1";
$result = mysql_query( $sql);
$cat_titles=array();
while( $row=mysql_fetch_assoc($result) )
{
$cat_titles[] = $row['id '];
// do stuff with other column
// data if we want
}
mysql_free_result( $result );
echo "<p>\n";
foreach($cat_titles as $v)
{
$cat_titles[]= $row['id'];
}
echo "</p>\n";
$cat_titles = implode(',',$cat_titles);
$cat_titles = substr($cat_titles,0,-2);
echo $cat_titles;
echo "select * from users where IN (".$cat_titles.")";
?>
A number of potential issues here:
You are not handling error conditions around you database access, so if you are having issue with your queries you would never know.
Your second select query doesn't specify a field in the WHERE clause, so it will never work
This section of code does absolutely nothing and is in fact where you problem likely lies.
foreach($cat_titles as $v)
{
$cat_titles[]= $row['id'];
}
Here $row['id'] won't have a value, so you are basically looping throguh your existing array and appending empty value to new indexes.
In all likelihood you could do this with a single query, it might help if you explain what you are actually trying to do.
You should not be using mysql_* functions. They are deprecated. Use mysqli or PDO instead.
i have a php question,
how can i get all the data from the same column from all row in a mysql table.
Table name : user
Here's my table structure:
name
url
How can i use php to get all the data in URL column from each row?
(P/S i have my connection to database established , just not sure the musql query for this)
Thanks and have a nice day.
This really is extremely basic stuff and you could have found this anywhere, it's even in the PHP manual. But alas, here you go.
$result = mysql_query("SELECT url FROM user");
while($row=mysql_fetch_assoc($result){
echo $row['url'].'\n';
}
Please read up on some basic stuff to avoid asking these kind of questions: http://www.freewebmasterhelp.com/tutorials/phpmysql
$query = "select url from user";
$result = mysql_query($query);
while( $row = mysql_fetch_assoc($result)){
echo $row['url'] . '<br>';
}