How to select and insert PHP Mysqli to the same table?
0
I am doing a one time sql select statement to see if a new record already exists in my database. If the record does not exist, then add the new record to the database.
That is all the logic to this code. Should be simple.
Error received: 1 Error description: Can't update table 'customer_info' in stored function/trigger because it is already used by statement which invoked this stored function/trigger.
Please rewrite my code so I can see how to write similar code in the future.
//What type of record are we searching for? eid or pid?
if($checkid) {
$t1 = "SELECT * FROM `customer_info` WHERE `e_id` LIKE '" .$row['ename'] . "';";
} else {
$t1 = "SELECT * FROM `customer_info` WHERE `p_id` LIKE '" .$row['pId'] . "';";
}
//Run the mysqli check to find any matching records
if($result = $mysqli->query($t1)){
printf("Select returned %d rows.n<br>", $result->num_rows);
}
//If a record was found, present it. Else, prepare to insert the new data.
if($info = mysqli_fetch_assoc($result)) {
print_r($info);
} else{
foreach($row as $key => $value) {
$row["$key"] = str_replace("'","", $value);
}
//Remove junk characters (please suggest a better way to do this)
unset($row['message']);
$values = "DEFAULT, '" . implode("', '",$row). "', '0'";
$values = str_replace("''","'",$values);
$values = str_replace("', ', '", "', '",$values);
$sql = "INSERT INTO `customer_info`(`id`, `transaction`, `price`,`ebay_id`, `paypal_id`, `eBook`, `item_num`, `quantity`, `first`, `last`, `street`, `country`, `state`, `zipcode`,`download`) VALUES ($values);";
if(!mysqli_query($mysqli, $sql)) {
echo ("1 Error description: " . mysqli_error($mysqli));
$mysqli->close(); die();
} else {
mysqli_query($mysqli, $sql);
$status = imap_setflag_full($inbox, $email_number, "\Seen", ST_UID);
}
}
php select mysqli insert stored-functions
asked Nov 22 '18 at 23:21
ChenelleChenelle
941515
add a comment |
0
I am doing a one time sql select statement to see if a new record already exists in my database. If the record does not exist, then add the new record to the database.
That is all the logic to this code. Should be simple.
Error received: 1 Error description: Can't update table 'customer_info' in stored function/trigger because it is already used by statement which invoked this stored function/trigger.
Please rewrite my code so I can see how to write similar code in the future.
//What type of record are we searching for? eid or pid?
if($checkid) {
$t1 = "SELECT * FROM `customer_info` WHERE `e_id` LIKE '" .$row['ename'] . "';";
} else {
$t1 = "SELECT * FROM `customer_info` WHERE `p_id` LIKE '" .$row['pId'] . "';";
}
//Run the mysqli check to find any matching records
if($result = $mysqli->query($t1)){
printf("Select returned %d rows.n<br>", $result->num_rows);
}
//If a record was found, present it. Else, prepare to insert the new data.
if($info = mysqli_fetch_assoc($result)) {
print_r($info);
} else{
foreach($row as $key => $value) {
$row["$key"] = str_replace("'","", $value);
}
//Remove junk characters (please suggest a better way to do this)
unset($row['message']);
$values = "DEFAULT, '" . implode("', '",$row). "', '0'";
$values = str_replace("''","'",$values);
$values = str_replace("', ', '", "', '",$values);
$sql = "INSERT INTO `customer_info`(`id`, `transaction`, `price`,`ebay_id`, `paypal_id`, `eBook`, `item_num`, `quantity`, `first`, `last`, `street`, `country`, `state`, `zipcode`,`download`) VALUES ($values);";
if(!mysqli_query($mysqli, $sql)) {
echo ("1 Error description: " . mysqli_error($mysqli));
$mysqli->close(); die();
} else {
mysqli_query($mysqli, $sql);
$status = imap_setflag_full($inbox, $email_number, "\Seen", ST_UID);
}
}
php select mysqli insert stored-functions
asked Nov 22 '18 at 23:21
ChenelleChenelle
941515
I believe that the error is self-explanatory.
– Funk Forty Niner
Nov 22 '18 at 23:39
But this isn't looping? I'm confused.
– Chenelle
Nov 22 '18 at 23:45
There appears to be a trigger/function happening somewhere in one of your databases/tables. If you're using phpmyadmin, you can check which processes are running.
– Funk Forty Niner
Nov 22 '18 at 23:46
add a comment |
0
0
0
I am doing a one time sql select statement to see if a new record already exists in my database. If the record does not exist, then add the new record to the database.
That is all the logic to this code. Should be simple.
Error received: 1 Error description: Can't update table 'customer_info' in stored function/trigger because it is already used by statement which invoked this stored function/trigger.
Please rewrite my code so I can see how to write similar code in the future.
//What type of record are we searching for? eid or pid?
if($checkid) {
$t1 = "SELECT * FROM `customer_info` WHERE `e_id` LIKE '" .$row['ename'] . "';";
} else {
$t1 = "SELECT * FROM `customer_info` WHERE `p_id` LIKE '" .$row['pId'] . "';";
}
//Run the mysqli check to find any matching records
if($result = $mysqli->query($t1)){
printf("Select returned %d rows.n<br>", $result->num_rows);
}
//If a record was found, present it. Else, prepare to insert the new data.
if($info = mysqli_fetch_assoc($result)) {
print_r($info);
} else{
foreach($row as $key => $value) {
$row["$key"] = str_replace("'","", $value);
}
//Remove junk characters (please suggest a better way to do this)
unset($row['message']);
$values = "DEFAULT, '" . implode("', '",$row). "', '0'";
$values = str_replace("''","'",$values);
$values = str_replace("', ', '", "', '",$values);
$sql = "INSERT INTO `customer_info`(`id`, `transaction`, `price`,`ebay_id`, `paypal_id`, `eBook`, `item_num`, `quantity`, `first`, `last`, `street`, `country`, `state`, `zipcode`,`download`) VALUES ($values);";
if(!mysqli_query($mysqli, $sql)) {
echo ("1 Error description: " . mysqli_error($mysqli));
$mysqli->close(); die();
} else {
mysqli_query($mysqli, $sql);
$status = imap_setflag_full($inbox, $email_number, "\Seen", ST_UID);
}
}
php select mysqli insert stored-functions
asked Nov 22 '18 at 23:21
ChenelleChenelle
941515
I am doing a one time sql select statement to see if a new record already exists in my database. If the record does not exist, then add the new record to the database.
That is all the logic to this code. Should be simple.
Error received: 1 Error description: Can't update table 'customer_info' in stored function/trigger because it is already used by statement which invoked this stored function/trigger.
Please rewrite my code so I can see how to write similar code in the future.
//What type of record are we searching for? eid or pid?
if($checkid) {
$t1 = "SELECT * FROM `customer_info` WHERE `e_id` LIKE '" .$row['ename'] . "';";
} else {
$t1 = "SELECT * FROM `customer_info` WHERE `p_id` LIKE '" .$row['pId'] . "';";
}
//Run the mysqli check to find any matching records
if($result = $mysqli->query($t1)){
printf("Select returned %d rows.n<br>", $result->num_rows);
}
//If a record was found, present it. Else, prepare to insert the new data.
if($info = mysqli_fetch_assoc($result)) {
print_r($info);
} else{
foreach($row as $key => $value) {
$row["$key"] = str_replace("'","", $value);
}
//Remove junk characters (please suggest a better way to do this)
unset($row['message']);
$values = "DEFAULT, '" . implode("', '",$row). "', '0'";
$values = str_replace("''","'",$values);
$values = str_replace("', ', '", "', '",$values);
$sql = "INSERT INTO `customer_info`(`id`, `transaction`, `price`,`ebay_id`, `paypal_id`, `eBook`, `item_num`, `quantity`, `first`, `last`, `street`, `country`, `state`, `zipcode`,`download`) VALUES ($values);";
if(!mysqli_query($mysqli, $sql)) {
echo ("1 Error description: " . mysqli_error($mysqli));
$mysqli->close(); die();
} else {
mysqli_query($mysqli, $sql);
$status = imap_setflag_full($inbox, $email_number, "\Seen", ST_UID);
}
}
php select mysqli insert stored-functions
php select mysqli insert stored-functions
asked Nov 22 '18 at 23:21
ChenelleChenelle
941515
asked Nov 22 '18 at 23:21
ChenelleChenelle
941515
asked Nov 22 '18 at 23:21
ChenelleChenelle
941515
asked Nov 22 '18 at 23:21
ChenelleChenelle
941515
asked Nov 22 '18 at 23:21
ChenelleChenelle
941515
941515
I believe that the error is self-explanatory.
– Funk Forty Niner
Nov 22 '18 at 23:39
But this isn't looping? I'm confused.
– Chenelle
Nov 22 '18 at 23:45
There appears to be a trigger/function happening somewhere in one of your databases/tables. If you're using phpmyadmin, you can check which processes are running.
– Funk Forty Niner
Nov 22 '18 at 23:46
add a comment |
I believe that the error is self-explanatory.
– Funk Forty Niner
Nov 22 '18 at 23:39
But this isn't looping? I'm confused.
– Chenelle
Nov 22 '18 at 23:45
There appears to be a trigger/function happening somewhere in one of your databases/tables. If you're using phpmyadmin, you can check which processes are running.
– Funk Forty Niner
Nov 22 '18 at 23:46
I believe that the error is self-explanatory.
– Funk Forty Niner
Nov 22 '18 at 23:39
I believe that the error is self-explanatory.
– Funk Forty Niner
Nov 22 '18 at 23:39
But this isn't looping? I'm confused.
– Chenelle
Nov 22 '18 at 23:45
But this isn't looping? I'm confused.
– Chenelle
Nov 22 '18 at 23:45
There appears to be a trigger/function happening somewhere in one of your databases/tables. If you're using phpmyadmin, you can check which processes are running.
– Funk Forty Niner
Nov 22 '18 at 23:46
There appears to be a trigger/function happening somewhere in one of your databases/tables. If you're using phpmyadmin, you can check which processes are running.
– Funk Forty Niner
Nov 22 '18 at 23:46
add a comment |
0
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I believe that the error is self-explanatory.
– Funk Forty Niner
Nov 22 '18 at 23:39
But this isn't looping? I'm confused.
– Chenelle
Nov 22 '18 at 23:45
There appears to be a trigger/function happening somewhere in one of your databases/tables. If you're using phpmyadmin, you can check which processes are running.
– Funk Forty Niner
Nov 22 '18 at 23:46