Question

我需要将concat函数的输出与where子句和like子句匹配。以下是我的查询

ng-model

我需要找到

的匹配结果

<div ng-app="myApp" >
<div  ng-controller="myCtrl">
     <input type="text" ng-model="name">
     <button ng-click="just()">submit</button>
</div>
<div  ng-controller="myCtrl2">
    <input type="text" ng-model="name2">
</div>
</div>

<script>
var app = angular.module('myApp', []);
app.controller('myCtrl', function($scope) {
    $scope.name="before";
    $scope.just = function()
    {
       $scope.name="after";

       $scope.name2="after";   /////  this is not working


    }
});
</script>

和

SELECT t.id, 
concat(trim(t.address1), ', ',t.zip, ' ',trim(t.city), ', ',c.countryName ) AS fullAddress 
FROM `User` `t` 
INNER JOIN Country c ON t.countryCode = c.countryCode

...

27 Avenue Pasteur, 14390 Cabourg, France

Answer 1

在查询中的最终t.zip之前，您在'，'中缺少空格。

应该是：

SELECT t.id, concat(trim(t.address1), ', ',t.zip, ' ',trim(t.city), ', ',c.countryName ) AS fullAddress FROM `User` `t` INNER JOIN Country c ON t.countryCode = c.countryCode WHERE (((address1 IS NOT NULL AND zip IS NOT NULL AND city IS NOT NULL AND t.countryCode IS NOT NULL) AND (concat( t.address1, ', ', t.zip, ' ', t.city, ', ', c.countryName ) regexp '^[0-9]+,? [^,]+, [0-9]+,? [^,]+, [a-zA-Z]+$')) AND (concat(' ',trim(t.address1), ', ',t.zip,' ', trim(t.city), ', ', c.countryName) like '%27 Avenue Pasteur, 143%'))

MYSQL concat导致where条件

1 个答案: