Here is a part of my php code:
foreach ($value->ahkam as $k => $v){
echo $v->id."\n";
//Save into db one hokm
$addHokm = "INSERT INTO qm_hokm (hokm_id, type, tooltip, line, x1, y1, x2, y2, radius, XOrigin, YOrigin, page_id)
VALUES ($v->id,$v->type,'tooltip',0,$v->x1,$v->y1,$v->x2,$v->y2,$v->r,$v->XOrigin,$v->YOrigin,$pageNumber)";
if(!mysqli_query($con, $addHokm))
echo "Failed to insert into db...".$v->id."\n";
}
In fact, I am fetching a json structure sent by an ajax request from a client.
I have many values in $value->ahkam but the problem is that only the first query is run and the others give me the error msg. Any help plz
UPDATE:
the result of echo is:
0
1
Failed to insert into db...1
2
Failed to insert into db...2
As you see, the hokm number 0 is added but not the others, I need to mention also that $pageNumer is a foeign key
The problem is in your foreign key, it must not be unique. Like that, you can add multiple entries for one page_id. I hope it is the correct answer:)
Based on your comments, it appears that your query is inserting a duplicate value for the page_id value, which appears to be set as a field that cannot have duplicate values. According to your query, you're using $pageNumber for that field, but I don't see it changing in your loop. You either need to get rid of the constraint preventing you from using the same value or make sure that $pageNumber has a value that isn't being used already.
Related
I have an external database that I am trying to access from within a Drupal page, I have successfully queried the database and output data to the page using fetchAssoc(), however this only returns the first row in the database. I would like to return all rows into an array for processing, so I'm attempting to use fetchAllAssoc(), this however results in an exception. The database has the following SQL fields:
id, model, manufacturer, url, date_modified
My test code is as follows:
<?php
db_set_active('product_db');
$query = db_select('product', 'p')->fields('p');
$sqlresults = $query->execute()->fetchAllAssoc('id');
foreach($sqlresults as $sqlresult)
{
printf($sqlresult);
}
db_set_active();
?>
I'm thinking that it is the key field 'id' that I am specifying with fetchAllAssoc() that is the problem, as fetchAssoc() prints values correctly. All documentation I have found seems to say that you pass a database field as the key but I have also passed a numeric value with no success.
Many thanks in advance for any advice, I'm sure I'm just missing something stupid.
I think it should work in this way, but within the foreach you want to print the $sqlresult variable as a string, but it is an object (it causes the error).
printf function needs a string as the first parameter, see:
http://php.net/manual/en/function.printf.php
Use for instance var_dump instead:
var_dump($sqlresult);
I can not get an SQL update statement to subtract a variable from a table value. Here is my code:
$_SESSION_Job101=mysql_fetch_array(mysql_query("SELECT * FROM job_101 WHERE job_101.username='$_SESSION_User'"));
mysql_query("UPDATE characters SET currenergy=currenergy-$_SESSION_Job101['ecost'] WHERE username='$_SESSION_User'");
$_SESSION_Job101 is a perfectly valid result, as I pull from it on another page; I even pull the 'ecost' on said page. I also update currenergy this way in another script, except I use the number 1 instead of the variable. So I've narrowed it down to that variable.
It wouldn't matter that $_SESSION_Job101 is the result from a second table (job_101), and that query is updating to the table characters, would it?
We don't have enough information, but since you don't perform ANY error handling or validation that SQL resultset is returned, it could be an error caused by issues such as:
no rows returned in first query
some other parsing issue not directly evident
I would propose that you use temporary strings and echo the actual SQL queries.
Continue by actually testing them with MYSQL (through workbench, queryviewer, or console) in order to see where and what the error is.
Also, it's not recommended to skip error checking and try to combine so many lines/steps into 2 lines.
Imagine the first query does not return any results for example...
Debugging:
$query1 = "SELECT * FROM job_101 WHERE job_101.username='$_SESSION_User'";
echo $query1."<br/>";
$_SESSION_Job101=mysql_fetch_array(mysql_query($query1 ));
$query2 = "UPDATE characters SET currenergy=currenergy-$_SESSION_Job101['ecost'] WHERE username='$_SESSION_User'";
echo $query2."<br/>";
mysql_query($query2);
Update
Based on your comment I suggest you try the following two options:
1) Add a space between the - and $_SESSION_Job101['ecost'].
2) If that doesn't work, change your string to:
mysql_query("UPDATE characters SET currenergy=currenergy-".$_SESSION_Job101['ecost']." WHERE username='".$_SESSION_User."'";`
I am developing an application in PHP for which I need to implement a big file handler.
Reading and writing the file is not a problem, but checking the content of the file is a problem.
I built a recursive function which checks whether or not a variable is already used in the same document.
private function val_id($id){
if(!isset($this->id)){
$this->id = array();
}
if(in_array($id, $this->id)){
return $this->val_id($id+1);
}else{
$this->id[] = $id;
return $id;
}
}
When in_array($id,$this->id) returns FALSE, the $id will be added to $this->id (array which contains all used ids) and returns a valid id.
When this returns TRUE, it returns the same function with parameter $id+1
Since we are talking about over 300000 records a time, PHP won't not to be able to store such big arrays. It seems to quit writing lines in the documents I generate when this array gets too big. But I don't receive any error messages like that.
Since the generated documents are SQL files with multiple rows INSERT another solution could be to check if the id already exists in the database. Can MySQL catch these exceptions and try these entries again with adding 1 to id? How?
How do you think I need to solve this problem?
Kind regards,
Wouter
make error messages to appear.
increase memory_limit
instead of values store the parameter in the key - so you'll be able to use isset($array[$this->id]) instead of in_array()
Use INSERT IGNORE to disable duplicate key check in mysql and remove your key check in php. Your statement could look like this.
INSERT IGNORE INTO tbl_name SET key1 = 1, col1 = 'value1'
If you want to add 1 to the id always you could use ON DUPLICATE KEY to increment your key by one:
INSERT INTO table (a,b,c) VALUES (1,2,3)
ON DUPLICATE KEY UPDATE c=c+1;
Why should 30.000 records be a problem? Each record in a standard PHP array takes 144 bytes, for 30.000 that would mean 4218,75 kByte. No big deal.
Otherwise, Your Common Sense's idea with the array-key is worth a thought, because it's faster.
I have to retrieve the history of a user and I have 4 tables whose data depend on each other.I can retrieve the data using loops,but I instead used the "where IN ()" clause and I implode the output of the previous query.However,if the list I provide to "where IN()" is empty it return an error.Is it that IN() cannot be empty?
When imploding an array for the IN clause, i do one of two things
1: Check if you even need to run the query at all
if(!empty($some_array)) {
//run mysql query
}
else {
// if you need to do something if the array is empty, such as error or set some defaults, do it here
}
2: A value in the array initiliser which is not ever in the database (for example, if im selecting based on a auto incrememnt id, i use zero as a default array value to stop any issues with empty data sets, as zero will never be in my id column).
$some_array = array(0);
You can add an empty value to the start, such as IN (0,your values here)
$blanknumber = $_POST["blankstartnumber"];
while ($blanknumber <= ($_POST["blankendnumber"] ))
{
echo "$blanknumber";
$blankid = $blanknumber;
$query = "INSERT INTO blank (Blank_ID) VALUES ('$blankid')";
mysql_query($query,$con);
$blanknumber++;
}
So the values are added into the database. Lets say if I have the starting number at 1 and ending at 5. It will all the those values, but it's still trying to add more into the database. I also tried adding an IF statement aswell. if ($blanknumber != $_POST["blankendnumber"])
12345 Error: Duplicate entry '5' for
key 'PRIMARY'
Make sure your $POST value is an integer; by default, I believe it will be cast as a string.
$_POST['varName'] = (int) $_POST['varName'];
edit:
$blanknumber = $_POST["blankstartnumber"];
while ($blanknumber <= ($_POST["blankendnumber"] ))
This should only execute once, since you're setting both comparison variables equal. Definitely 2x check your code.
The database error indicates that Blank_ID is your primary key for that table, and you'd already inserted a 5 into the row. A primary key's values can exist only once in the entire table - duplicates are forbidden (if they were allowed, it wouldn't be a primary key anymore).
If your while loop isn't ending, I'd suggest dumping out both the blankendnumber and blankstartnumber before the loop starts, making sure you've got the right values in there.
It looks like it's actually functioning properly, but you might not have tidy'ed up your db table prior to running. If your output was:
123455 Error: Duplicate entry '5'...
Then, you'd have a programming error, as 5 is getting run twice. Instead, I think you already have data in the blank table that causes a conflict.
Edit: to automatically have MySQL handle the duplicate key error gracefully, you can use the ON DUPLICATE KEY clause to update the row.
INSERT INTO blank (Blank_ID) VALUES (5) ON DUPLICATE KEY UPDATE mod_date = NOW();