он выводит соответственно произведение необходимых результатов,
есть связь таблиц...как его переписать при помощи LEFT JOIN, учитывая связи???
Возможно ли это???
Код
$sql = "SELECT $table_columns_list FROM $table_unique_list WHERE 1=2";
for ($i=0; $i < count($table_columns); $i++) {
$sql .= ' or ' . $table_columns[$i] . " LIKE '%$search%'";
for ($i=0; $i < count($table_columns); $i++) {
$sql .= ' or ' . $table_columns[$i] . " LIKE '%$search%'";
Код
$table_unique_list = certificates, general, organization, statuses, categories, products
$table_columns_list = certificates.serial_number, certificates.action_character,
certificates.delivery_date, certificates.termination_date, certificates.assurance_level,
general.date_of_publication, general.date_of_renovation, organization.organization_name,
statuses.status_name, organization.city, organization.street, organization.house,
organization.housing, organization.phone_number, categories.category_name,
products.long_title
$table_columns[$i] = certificates.serial_number, certificates.action_character,
certificates.delivery_date, certificates.termination_date, certificates.assurance_level,
general.date_of_publication, general.date_of_renovation, organization.organization_name,
statuses.status_name, organization.city, organization.street, organization.house,
organization.housing, organization.phone_number, categories.category_name,
products.long_title
$table_columns_list = certificates.serial_number, certificates.action_character,
certificates.delivery_date, certificates.termination_date, certificates.assurance_level,
general.date_of_publication, general.date_of_renovation, organization.organization_name,
statuses.status_name, organization.city, organization.street, organization.house,
organization.housing, organization.phone_number, categories.category_name,
products.long_title
$table_columns[$i] = certificates.serial_number, certificates.action_character,
certificates.delivery_date, certificates.termination_date, certificates.assurance_level,
general.date_of_publication, general.date_of_renovation, organization.organization_name,
statuses.status_name, organization.city, organization.street, organization.house,
organization.housing, organization.phone_number, categories.category_name,
products.long_title
А связи выглядят следующим образом:
Код
organization => general <= categories => products
и products => general
и products => general
statuses, certificates - не связаны с остальными таблицами из запросаю.
Я понимаю, что statuses, certificates надо будет связать с остальными. А с какими лучше???
Помогите хотя бы разобраться на примере связанных таблиц. Очень надо! Пожалуйста!!!
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