如何将字符串连接到结果集 [英] How to I concatenate a string to a result set

问题描述

 var query =SELECT * FROM measurements WHERE project_type =？AND project_id =？AND subproject =？LIMIT 1; 
 
 
 
 $ cordovaSQLite.execute（db，query，['SMBH'，$ scope.proj1，'钻进和完成5英寸直径钻孔底座'）。然后（函数（结果）{
 
 $ scope.results = []; 
 
 if（result.rows.length> 0）{
 
 for（var i = 0; i< result.rows.length; i ++）{
 // alert（items exists）; 
 sub [add] = result.rows.item（i ）.address; 
 sub [award] = result.rows.item（i）.date_awarded; 
 sub [rep] = result.rows.item（i）.reporter_id; 
 sub [contr] = result.rows.item（i）.contractor_id; 
 for（var m = 1; m< = 49; m ++）{
 var l =l; 
 sub [l+ m] = result.rows.item（i）.l1; // FETCHES结果
 sub [l+ m] = result.rows.item（i） + l + m; //没有结果结果
 
 sub [b+ m] = result.rows.item（i）.b1; //获得结果
 sub [d+ m] = result.rows.item（i）.d + m; // DOES Not FETCH结果
 sub [num+ m] = result.rows.item（i ）.num1; // FETCHES结果
 sub [cr+ m] = result.rows.item（i）.certified_rate1; // FETCHES RES ULT 

我尝试过：



sub [d+ m] = result.rows.item（i）.d + m;

sub [d+ m] = result.rows.item（i）。 d+ m;

sub [d+ m] = result.rows.item（i）.d + [m];

sub [d + m] = result.rows.item（i）+。+ d + [m];

解决方案

cordovaSQLite.execute（db，query，['SMBH '，

scope.proj1，'钻进和完成5英寸直径钻孔底座']。然后（功能（结果）{

范围。结果= [];

if（result.rows.length> 0）{

for（var i = 0; i< result.rows.length; i ++）{
// alert（items exists）;
sub [add] = result.rows.item（i）.address;
sub [award] = result.rows.item（i）.date_awarded;
sub [rep] = result.rows.item（i）.reporter_id;
sub [contr] = result.rows.item（i）.contractor_id;
for（var m = 1; m< = 49; m ++）{
var l =l;
sub [l+ m] = result.rows.item（i）.l1; // FETCHES结果
sub [l+ m] = result.rows.item（i） + l + m; //没有结果结果

sub [b+ m] = result.rows.item（i）.b1; //获得结果
sub [d+ m] = result.rows.item（i）.d + m; // DOES Not FETCH结果
sub [num+ m] = result.rows.item（i ）.num1; // FETCHES结果
sub [cr+ m] = result.rows.item（i）.certified_rate1; //获得结果

我尝试过：



sub [d+ m] =结果。 rows.item（i）.d + m;

sub [d+ m] = result.rows.item（i）。d+ m;

sub [d+ m] = result.rows.item（i）.d + [m];

sub [d+ m] = result.rows.item（i）+ + d + [M];

var query=" SELECT * FROM measurements WHERE project_type=? AND project_id= ? AND subproject=? LIMIT 1";



$cordovaSQLite.execute(db, query, ['SMBH',$scope.proj1,'DRILLING AND COMPLETION OF 5 INCH DIAMETER BOREHOLE BASEMENT']).then(function(result) {

$scope.results=[]; 

if(result.rows.length >0){

for(var i=0; i<result.rows.length; i++){
   //alert("items exists");
sub["add"] = result.rows.item(i).address;
sub["award"] = result.rows.item(i).date_awarded;
sub["rep"] = result.rows.item(i).reporter_id;
sub["contr"]= result.rows.item(i).contractor_id;
for(var m=1; m<=49;m++){
var l="l";
sub["l"+m] = result.rows.item(i).l1;//FETCHES THE RESULT
sub["l"+m] = result.rows.item(i)+l+m;//DOESN'T FETCH THE RESULT

sub["b"+m] = result.rows.item(i).b1;//FETCHES THE RESULT
sub["d"+m] = result.rows.item(i).d+m;//DOESN't FETCH THE RESULT
sub["num"+m] = result.rows.item(i).num1;//FETCHES THE RESULT
sub["cr"+m] = result.rows.item(i).certified_rate1;//FETCHES THE RESULT

What I have tried:

sub["d"+m] = result.rows.item(i).d+m;
sub["d"+m] = result.rows.item(i)."d"+m;
sub["d"+m] = result.rows.item(i).d+[m];
sub["d"+m] = result.rows.item(i)+"."+d+[m];

解决方案

cordovaSQLite.execute(db, query, ['SMBH',

scope.proj1,'DRILLING AND COMPLETION OF 5 INCH DIAMETER BOREHOLE BASEMENT']).then(function(result) {

scope.results=[]; if(result.rows.length >0){ for(var i=0; i<result.rows.length; i++){ //alert("items exists"); sub["add"] = result.rows.item(i).address; sub["award"] = result.rows.item(i).date_awarded; sub["rep"] = result.rows.item(i).reporter_id; sub["contr"]= result.rows.item(i).contractor_id; for(var m=1; m<=49;m++){ var l="l"; sub["l"+m] = result.rows.item(i).l1;//FETCHES THE RESULT sub["l"+m] = result.rows.item(i)+l+m;//DOESN'T FETCH THE RESULT sub["b"+m] = result.rows.item(i).b1;//FETCHES THE RESULT sub["d"+m] = result.rows.item(i).d+m;//DOESN't FETCH THE RESULT sub["num"+m] = result.rows.item(i).num1;//FETCHES THE RESULT sub["cr"+m] = result.rows.item(i).certified_rate1;//FETCHES THE RESULT

What I have tried:

sub["d"+m] = result.rows.item(i).d+m;
sub["d"+m] = result.rows.item(i)."d"+m;
sub["d"+m] = result.rows.item(i).d+[m];
sub["d"+m] = result.rows.item(i)+"."+d+[m];

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