Fix table name and avoid double id in query.
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fe02ebe2fb
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@ -60,14 +60,16 @@ for ($i = 97; $i <= 122 ; $i++) {
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if ($result->num_rows > 0) {
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if ($result->num_rows > 0) {
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$query = "DELETE FROM " . $mysql['database'] . " WHERE id = $sender_id ;";
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$query = "DELETE FROM data WHERE id = $sender_id ;";
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$conn->query($query);
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$conn->query($query);
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}
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}
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$query = "INSERT INTO data values( $sender_id";
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$query = "INSERT INTO data values( $sender_id";
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foreach ($alphabet as $letter=>$num){
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foreach ($alphabet as $letter=>$num){
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$query .= "," . $num;
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if ($letter != "id") {
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$query .= "," . $num;
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}
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}
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}
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$query .= ");";
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$query .= ");";
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