i am trying to insert into multi table after select query return 0 (not found raws) select query working and insert query never done when submite "displayid" and there is no any syntax error
code:
<?php
if ($_POST["displayid"] == TRUE) {
$sqlid = "SELECT * FROM doc1 WHERE idnum ='$pidnum' AND stats='$ok'";
$result = mysqli_query($conn, $sqlid);
if (mysqli_num_rows($result) > 0) {
$sqlup = "UPDATE doc1 SET m_phone='$pm_phone', seen='$dataseen' WHERE idnum ='$pidnum'";
mysqli_query($conn, $sqlup);
$found = 1;
} else {
$found = 0;
$sqlfail = "INSERT INTO fail(fname,lname,tname,funame,idnum,m_phone,reg_date)
VALUES ('$pfname','$plname','$ptname','$pfuname','$pidnum','$pm_phone','$todaydate')";
$conn->query($sqlfail)
}
}
?>
Use this code:
$sqlfail = "INSERT INTO fail(fname,lname,tname,funame,idnum,m_phone,reg_date)
VALUES ('".$pfname."','".$plname."','".$ptname."','".$pfuname."','".$pidnum."','".$pm_phone."','".$todaydate."')";
make similar changes for update command as well
you actually have one error
$conn->query($sqlfail)
should be
$conn->query($sqlfail);
AND stats='$ok'";
i can't see a variable with this name i think you mean AND stats='ok'";
Related
I have this php code:
$query = $database->query("SELECT EXISTS(SELECT * FROM contacts WHERE contact_id = '$contactID')";
if($query == 0){
echo "not registered";
}elseif($query == 1){
echo "registered"
}
If I'm not wrong, the query is suppose to return 0 or 1 and it works in my SQLite manager. What is the correct way on getting that value in Php and use it in IF ELSE statement?
If you only need a single value, you can use querySingle:
$result = $database->querySingle("SELECT EXISTS(SELECT * FROM contacts WHERE contact_id = '$contactID'");
Otherwise, with normal queries, the result returned by ->query isn't actually the data itself, but an identifier you would use to get data from the database:
$results = $db->query('SELECT bar FROM foo');
while ($row = $results->fetchArray()) {
var_dump($row);
}
I have a problem when trying to update table after checking row. Not sure if the "if" statement is wrong, however I'm not quite sure, why the UPDATE sql is returning this error. I wouldn't be suprised if INSERT did that.
Here's part of code:
$sql = "SELECT user_id FROM players WHERE user_id = '$id'";
$result = $connect->query($sql);
if($result->num_rows > 0)
{
$sql = "UPDATE players SET user_id = '$Player->user_id', display_name = '$Player->display_name', attackPower = '$Player->attackPower]', defensePower = '$Player->defensePower'";
if($connect->query($sql) === TRUE)
{
echo 'Table has been successfully updated.';
}else{
echo 'There has been a problem with updating the "players" table. <br>Error: '.$connect->error;
}
}else{
$sql = "INSERT INTO players(user_id, display_name, attackPower, defensePower) VALUES('$Player->user_id', '$Player->display_name', '$Player->attackPower', '$Player->defensePower')";
if($connect->query($sql) === TRUE)
{
echo'Table has been successfully migrated.';
}else{
echo'Table migration has failed.';
}
}
$connect->close();
INSERTing is working just fine. I would appreciate any advice. Thanks.
Your update query should look like:
$sql = "UPDATE `players` SET `display_name` = '{$Player->display_name}',
`attackPower` = '{$Player->attackPower}', `defensePower` = '{$Player->defensePower'}
WHERE `user_id` = '{$Player->user_id}'";
It cause an error because Identity columns are not updateable.
You can update every columns except them:
$sql = "UPDATE players SET display_name = '$Player->display_name', attackPower = '$Player->attackPower]', defensePower = '$Player->defensePower'";
As #aynber and #Julqas said, problem was my sql was missing WHERE condition. Thanks for help.
$sql = "SELECT count(id),soft_name from table_name GROUP BY soft_name";
$d = mysqli_fetch_assoc(mysqli_query($db_name, $sql));
$c = array_shift($d);
The result is always 2, but the database contains more than 3000 items. What could be the problem?
Records are not pulled the way you are pulling for that you have to use similar to following code:
if($result = mysqli_query($db_name, $sql)){
while($d = mysqli_fetch_assoc($result){
echo $d['count'];
}
}
More reference here.
I'm trying to check if the leaguename name already exists if it does then check if it is higher than the existing score. if its higher then replace the quizscore else insert the leaguename and quizscore into the table league_quiz.
The code seem to insert the values in my table, but it do not seem update if the name is equal to previous and the score is higher?
I get this error:
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, string given in
My code:
<?php
$name = (string)$_POST['name'];
$score = (string) $_POST['score'];
$link = mysqli_connect("mysql12.gigahost.dk","username","password","dirts_mysql");
$query = "SELECT leaguename, quizscore FROM league_quiz WHERE leaguename = '$name' AND quizscore < '$score'";
if(mysqli_num_rows($query)) {
mysqli_query($link,"UPDATE league_quiz SET quizscore = '$score'");
} else {
mysqli_query($link,"INSERT INTO league_quiz (leaguename, quizscore) VALUES ('$name', '$score')") or die(mysqli_error($link));
}
?>
mysqli_num_rows expects the result of a query and not the string of the query itself.
Firstly you must execute your query :
$result = mysqli_query($link,$query);
and then you can count the number of rows
$numRows = mysqli_num_rows($result);
Between those two:
$query = "SELECT leaguename, quizscore FROM league_quiz WHERE leaguename = '$name' AND quizscore < '$score'";
if(mysqli_num_rows($query)) {
Have you tried to write this code?
$result = mysqli_query($query);
and then use mysqli_num_rows on the result of the query?
Try the followowing:
$result = mysqli_query($query);
if (mysqli_num_rows($result)) {
...
I'm working with a file and I'm attempting to do multiple select statements one after another and insert some values. So far the insert and the select I've got working together but when attempting to get the last SELECT to work I get no value. Checking the SQL query in workbench and everything works fine. Here's the code:
$id = "SELECT idaccount FROM `animator`.`account` WHERE email = '$Email'";
$result = mysqli_query($dbc, $id) or die("Error: ".mysqli_error($dbc));
while($row = mysqli_fetch_array($result))
{
echo $row[0];
$insert_into_user = "INSERT INTO `animator`.`user` (idaccount) VALUES ('$row[0]')";
}
$select_userid = "SELECT iduser FROM `animator`.`user` WHERE iduser = '$row[0]'";
$results = mysqli_query($dbc, $select_userid) or die("Error: ".mysqli_error($dbc));
while($rows = mysqli_fetch_array($results))
{
echo $rows[0];
}
I do not want to use $mysqli->multi_query because of previous problems I ran into. Any suggestions? And yes I know the naming conventions are close naming... They will be changed shortly.
Your code makes no sense. You repeatedly build/re-build the $insert_int-User query, and then NEVER actually execute the query. The $select_userid query will use only the LAST retrieved $row[0] value from the first query. Since that last "row" will be a boolean FALSE to signify that no more data is available $row[0] will actually be trying to de-reference that boolean FALSE as an array.
Since you're effectively only doing 2 select queries (or at least trying to), why not re-write as a single two-value joined query?
SELECT iduser, idaccount
FROM account
LEFT JOIN user ON user.iduser=account.idaccount
WHERE email='$Email';
I'm not sure what you're trying to do in your code exactly but that a look at this...
// create select statement to get all accounts where email=$Email from animator.account
$id_query = "SELECT idaccount FROM animator.account WHERE email = '$Email'";
echo $id_query."\n";
// run select statement for email=$mail
$select_results = mysqli_query($dbc, $id_query) or die("Error: ".mysqli_error($dbc));
// if we got some rows back from the database...
if ($select_results!==false)
{
$row_count = 0;
// loop through all results
while($row = mysqli_fetch_array($result))
{
$idaccount = $row[0];
echo "\n\n-- Row #$row_count --------------------------------------------\n";
echo $idaccount."\n";
// create insert statement for this idaccount
$insert_into_user = "INSERT INTO animator.user (idaccount) VALUES ('$idaccount')";
echo $insert_into_user."\n";
// run insert statement for this idaccount
$insert_results = mysqli_query($dbc, $insert_into_user) or die("Error: ".mysqli_error($dbc));
// if our insert statement worked...
if ($insert_results!==false)
{
// Returns the auto generated id used in the last query
$last_inisert_id = mysqli_insert_id($dbc);
echo $last_inisert_id."\n";
}
else
{
echo "insert statement did not work.\n";
}
$row_count++;
}
}
// we didn't get any rows back from the DB for email=$Email
else
{
echo "select query returned no results...? \n";
}