I'm trying to add an autocomplete tokenizer script to some form fields and one issue I'm having is if a person saves multiple values for the field the autocomplete suggestions come back with all of his values as one long value instead of them being single values delimited by the comma. I first tried to simply explode the value but it doesn't format it correctly in the JSON encode.
Here is my PHP file:
//connection information
$host = "localhost";
$user = "myuser";
$password = "mypass";
$database = "mydb";
$param = ($_GET["term"]);
//make connection
$server = mysql_connect($host, $user, $password);
$connection = mysql_select_db($database, $server);
//query the database
$query = mysql_query("SELECT cb_activities FROM jos_comprofiler WHERE cb_activities REGEXP '^$param'");
//build array of results
for ($x = 0, $numrows = mysql_num_rows($query); $x < $numrows; $x++) {
$row = mysql_fetch_assoc($query);
$activities[$x] = array(cb_activitiesterm => $row[cb_activities]);
}
//echo JSON to page
$response = $_GET["callback"] . "(" . json_encode($activities) . ")";
echo $response;
mysql_close($server);
This gives the output like this:
[{"cb_activities":"Kicking Cats,"},{"cb_activities":"baseball,hockey,"}]
but I need it to output like this:
[{"cb_activities":"Kicking Cats,"},{"cb_activities":"baseball,"},"cb_activities":"hockey,"}]
I also need to find a way to prevent duplicate entries from populating. For instance, the way it is now, say 10 people all have kicking cats selected as a value, it will display 10 times in the autocomplete suggestions.
How do I set this up to correctly delimit at the commas and then weed out duplicate values?
NM the duplicate issue, I just added select distinct instead of just select, this json thing has me overcomplicating things now lol. Now if I can just figure out how to delimit properly at the comma all will be good.
Related
I have a database field that contains emails seperated by a comma as below
email1#domain.com,email2#domain.com, etc etc
I am using PHP Mailer and have found several articles on here explaining how to loop through each email to add each email to the emailto address
I'm clearly doing something wrong as my code below only sends to one email
$sql="SELECT Email FROM Managers WHERE Company = '$name';";
$result = mysqli_query($db,$sql);
while ($row = mysqli_fetch_assoc($result));
$address = explode (',' , $row['Email']);
foreach ($address as $key => $v) {$mail->addAddress($key);}
When I output the result to check the output I just see all the emails as one string with no commas
Any advice would be most appreciated as from what I have read on here this should work.
UPDATE: I now have this working as below. Thanks for your help:
$sql="SELECT Email FROM Managers WHERE Company = '$name';";
$result = mysqli_query($db,$sql);
while ($row = mysqli_fetch_array($result)) {
$address = $row['Email'];
$address2 = explode (',',$address);
foreach ($address2 as $v) {$mail->addAddress($v);}
Always check what your code is really doing, don't assume it's doing what you think it is.
There is a simple bug in your code:
foreach ($address as $key => $v) {$mail->addAddress($key);}
Should be:
foreach ($address as $v) {$mail->addAddress($v);}
otherwise you're submitting the column name as the email address.
I'm using a json-parser in Xcode to fetch a table from phpmyadmin. The parser gets (or should get) the json-formated document via a php-file uploaded on my ftp-server. The file is successfully parsed but it doesn't recognize any objects. I think this is because there are multiple arrays in the json-document.
When there's only one entry the document looks like this:
[{"id":"1","Name":"Eric","Message":"first from web"}]
but when i add an entry it looks like this:
[{"id":"1","Name":"Eric","Message":"first from web"}]
[{"id":"1","Name":"Eric","Message":"first from web"},{"id":"6","Name":"Claes","Message":"Hurrburr"}]
As you can see the old array (containing only the single entry) is still there in the second array with both entries.
I suspect the problem is that the old arrays are still there when i update the database because when i tried parsing the json document with only one entry (only one array) it worked.
So my question is first if there´s something i missed in my code, or if you know why the old arrays are still there when I update the database or how to remove all previous arrays when the document is updating.
Here is my .php-file:
<?php
$username = "perhaps not sharing this information";
$password = "or this";
$database = "nah";
mysql_connect("the server url",$username, $password);
#mysql_select_db($database) or die("Error here");
$query = "SELECT * FROM debug_db";
$result = mysql_query($query) or die(mysql_error());
$num = mysql_numrows($result);
mysql_close();
$rows = array();
while($r = mysql_fetch_assoc($result))
{
$rows[] = $r;
echo json_encode($rows);
}
?>
And check out the json-document at: http://app.levinnovation.se/getjson.php
Thank you!
Using PDO I have built a succinct object for retrieving rows from a database as a PHP object with the first column value being the name and the second column value being the desired value.
$sql = "SELECT * FROM `site`"; $site = array();
foreach($sodb->query($sql) as $sitefield){
$site[$sitefield['name']] = $sitefield['value'];
}
I now want to apply it to a function with 2 parameters, the first containing the table and the second containing any where clauses to then produce the same result.
function select($table,$condition){
$sql = "SELECT * FROM `$table`";
if($condition){
$sql .= " WHERE $condition";
}
foreach($sodb->query($sql) as $field){
return $table[$field['name']] = $field['value'];
}
}
The idea that this could be called something like this:
<?php select("options","class = 'apples'");?>
and then be used on page in the same format as the first method.
<?php echo $option['green'];?>
Giving me the value of the column named value that is in the same row as the value called 'green' in the column named field.
The problem of course is that the function will not return the foreach data like that. That is that this bit:
foreach($sodb->query($sql) as $field){
return $table[$field['name']] = $field['value'];
}
cannot return data like that.
Is there a way to make it?
Well, this:
$sql = "SELECT * FROM `site`"; $site = array();
foreach($sodb->query($sql) as $sitefield){
$site[$sitefield['name']] = $sitefield['value'];
}
Can easily become this:
$sql = "SELECT * FROM `site`";
$site = array();
foreach( $sodb->query($sql) as $row )
{
$site[] = $row;
}
print_r($site);
// or, where 0 is the index you want, etc.
echo $site[0]['name'];
So, you should be able to get a map of all of your columns into the multidimensional array $site.
Also, don't forget to sanitize your inputs before you dump them right into that query. One of the benefits of PDO is using placeholders to protect yourself from malicious users.
Based on this question How to insert array into mysql using PDO and bindParam?
I'm trying to insert values of an array into mysql via PDO.
I'm having a hard time of it, because I keep getting the following error.
SQLSTATE[HY093]: Invalid parameter number: number of bound variables does not match number of tokens
for this line $stmt->execute();
I'm guessing the problem has something to do with this line
$stmt->bindParam(':val$count', $val,PDO::PARAM_STR); Specifically 'val$count', but I'm not sure exactly what is going wrong.
QUESTION: What am I doing wrong? How can I fix this?
Anyway here is the code I'm using along with the sample array.
$lastInsertValue=87;
$qid[0][0]=1;
$qid[0][1]=1;
$qid[1][0]=2;
$qid[1][1]="null";
$qid[2][0]=3;
$qid[2][1]=0;
$array_count = count($qid);
if (isset($lastInsertValue))
{
try
{
$dbh = new PDO("mysql:host=$hostname;dbname=$dbname", $username, $password);
$stqid=array();
$a=0;
for ($i=0; $i<$array_count; $i++)
{
$stqid[$a]=$lastInsertValue;
$a++;
$stqid[$a]=$qid[$i][0];
$a++;
$stqid[$a]=$qid[$i][1];
$a++;
}
$sql = "INSERT INTO qresults (instance, qid, result) VALUES ( :val0, :val1, :val2)";
$count = 0;
$stmt = $dbh->prepare($sql);
foreach ($stqid as $val)
{
$stmt->bindParam(':val$count', $val,PDO::PARAM_STR);
$count++;
}
$stmt->execute();
$dbh = null;
}
catch(PDOException $e)
{
echo $e->getMessage();
}
}
Yikes, so many issues.
Your array building is so verbose. Try this
$stqid = array();
foreach ($qid as $qidArr) {
$stqid[] = $lastInsertValue; // no idea why you repeat this
$stqid[] = $qidArr[0];
$stqid[] = $qidArr[1];
}
Use positional placeholders if you're simply relying on number of arguments
$sql = 'INSERT INTO ... VALUES (?, ?, ?)';
bindParam uses references which you are overwriting with each loop iteration. You would want to use bindValue() instead
Your query only has 3 placeholders but your $stqid array has 9 items. This is the source of your error.
You have single quotes around a variable, which will be treated as the variable name $count (not the value), try concatenating the variable to the string. Give this a try:
$stmt->bindParam(':val' . $count, $val,PDO::PARAM_STR);
for ($i=0; $i<$array_count; $i++)
{
$stqid[$a]=$lastInsertValue;
$a++;
$stqid[$a]=$qid[$i][0];
$a++;
$stqid[$a]=$qid[$i][1];
$a++;
}
so in case $i = 2, it will add $stqid[6], $stqid[7], $stqid[8] so
foreach ($stqid as $val)
{
$stmt->bindParam(':val$count', $val,PDO::PARAM_STR);
$count++;
}
will give you :val0 to :val8
In your query you have only :val0 to :val2.
Also having multiple values in one field in database is bad. Don't do it. Try to redesign your DB differently
EDIT: bad math in the morning... sorry
I'm using PHP to make a very specific sql query. For example sake, I have the user's ID number, but I need their name. So I do a sql query from that table with the ID number in order to return the name.
$result = mysql_query("SELECT name FROM users WHERE userID=$thisuserid",$db);
Now I want to use that. What's the most succinct way to go about making that result into a variable ths I can use?
edit:
I'm hoping that this is not the answer:
$rowCheck = mysql_num_rows($result);
if ($rowCheck > '0') {
while ($row = mysql_fetch_assoc($result)){
foreach ($row as $val){
$username = $val;
}
}
}
I have used something like this to keep it short in the past:
list($name) = mysql_fetch_row(mysql_query("SELECT name FROM users WHERE userID=$thisuserid",$db));
echo $name;
In my opinion, the best way to fetch any SQL result is through mysql_fetch_assoc(). To use it, you would do something like this:
$result = mysql_query("SELECT name FROM users WHERE userID=$thisuserid",$db);
while ($row = mysql_fetch_assoc($result)) {
echo $row['name']; // You get an array with each column returned from your query.
}
Still, MySQL extension has been replaced for MySQLi, which is acknowledged to be faster and more practical. It has both OOP and structural bindings, and takes more into account your server settings.
$result = mysql_query("SELECT name FROM users WHERE userID=$thisuserid",$db);
if (!$result) {
echo 'Could not run query: ' . mysql_error();
exit;
}
$name = mysql_fetch_row($result)[0];
You should use MySQLi as bellow:
$db = new MySQLi($host,$user,$pass,$db);
$query = $db->query('SELECT name FROM users WHERE userID='.$thisuserid);
$result = $query->fetch_object();
echo $result->name;
If you use SELECT * so you also can access via $result->{field_name}