我想将 PostgreSQL 中的行转换为列。我希望所有变量都针对其各自的 id。但它不起作用。
预期输出:
myvar desc fname lname sdate edate id
title1 desc1 cina jhon 1483920000000 1484524800000 14
title2 desc2 jhon lname2 1483920000000 1483910000000 16
title3 desc3 diesel zier 1483920000000 1484524800000 17
SELECT * FROM crosstab(
'SELECT name, value, id FROM test ORDER BY id') AS (
"myVar" text, "desc" text, "fname" text, "lname" text,"sdate" text,"edate" text, "value" text ,"containerid" bigint);
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错误:错误:返回类型无效 SQL 状态:42601 详细信息:SQL rowid 数据类型与返回 rowid 数据类型不匹配。
我正在尝试打开jquery对话框,但它没有打开我也使用jquery datepicker.当我删除日期选择器的两个jquery脚本时,对话框打开,否则它无法以任何方式工作.任何人都建议我使用任何jquery脚本的对话框,可以使用这些datepicker脚本.
javascript datepicker代码:
<script src="./jquery.js"></script>
<script src="./jquery.datetimepicker.js"></script>
<script>/*
window.onerror = function(errorMsg) {
$('#console').html($('#console').html()+'<br>'+errorMsg)
}*/
$('#datetimepicker').datetimepicker({
dayOfWeekStart : 1,
lang:'en',
disabledDates:['1986/01/08','1986/01/09','1986/01/10'],
startDate: '1986/01/05'
});
$('#datetimepicker').datetimepicker({value:'2015/04/15 05:03',step:10});
$('.some_class').datetimepicker();
$('#default_datetimepicker').datetimepicker({
formatTime:'H:i',
formatDate:'d.m.Y',
//defaultDate:'8.12.1986', // it's my birthday
defaultDate:'+03.01.1970', // it's my birthday
defaultTime:'10:00',
timepickerScrollbar:false
});
$('#datetimepicker10').datetimepicker({
step:5,
inline:true
});
$('.datetimepicker_mask').datetimepicker({
mask:'9999/19/39 29:59'
});
$('#datetimepicker1').datetimepicker({
datepicker:false,
format:'H:i',
step:5
});
$('#datetimepicker2').datetimepicker({
yearOffset:222,
lang:'ch',
timepicker:false,
format:'d/m/Y',
formatDate:'Y/m/d',
minDate:'-1970/01/02', // yesterday is minimum date
maxDate:'+1970/01/02' // and tommorow is maximum date calendar
});
$('#datetimepicker3').datetimepicker({
inline:true
}); …Run Code Online (Sandbox Code Playgroud)