データベースにデータを挿入するフォームがあります。フィールドの 1 つはweDate
、
$weDate=date('Y-m-d',strtotime('Friday'))
以下を使用する場合:
$sql = mysql_query("SELECT ID as 'DB_ID',
partnumber as 'Part_Number',
pndesc as 'Part Number Description',
name as 'Name',
reason as 'Reason',
comment as 'Comments',
date as 'Date',
time as 'Time',
weDate as 'Weekend Date'
FROM $table
WHERE weDate = '2013-05-03'");
結果が返ってきます。
私がしようとすると:
$sql = mysql_query("SELECT ID as 'DB_ID',
partnumber as 'Part_Number',
pndesc as 'Part Number Description',
name as 'Name',
reason as 'Reason',
comment as 'Comments',
date as 'Date',
time as 'Time',
weDate as 'Weekend Date'
FROM $table
WHERE weDate = '$weDate2'");
何も返ってきません。なんで?
PHP 変数は次のとおりです。
$weDate2 = date('Y-m-d', strtotime('-1 weeks Friday')