SQL錯誤在PHP返回的查詢與聯接

我試圖通過執行以下查詢的PHP應用程序來填充表：SQL錯誤在PHP返回的查詢與聯接

$sql3 = "SELECT distinct(`t1.testName`), `t2.comments AS C1` from `sample AS t1` left join `sample AS t2` ON `t1.testName`= `t2.testName` where `t1.buildNumber`= 181 and `t2.buildNumber`= 180 and `t1.errorStackTrace` is not null"; 

$result3 = mysqli_query($dbconnect,$sql3); 

if(!mysqli_query($dbconnect, $sql3)){ 
    printf("error message: %s\n",mysqli_error($dbconnect)); 
}

我看到下面的錯誤返回：

error message: Table 'testdata.sample as t1' doesn't exist

我已經嘗試了很多來解決這個問題，但不能。在MySQL上運行時查詢運行良好。任何幫助，將不勝感激。謝謝

來源

2016-07-28 Saurabh

使用適當的反引號或從示例AS t1中刪除'$ sql3 =「SELECT distinct（t1.testName），t2.comments AS C1 left left join sample AS t2 ON t1.testName = t2.testName其中t1.buildNumber = 181和t2.buildNumber = 180和t1 .errorStackTrace不爲空「; ' –

您必須使用反引號唯一的角落找尋表名或列名不包括別名：

$sql3 = "SELECT distinct(`t1`.`testName`), `t2`.`comments` AS C1 from `sample` AS ` left join `sample` AS t2 ON t1.testName= t2.testName where t1.buildNumber= 181 and t2.buildNumber= 180 and t1.errorStackTrace is not null";

來源

2016-07-28 05:22:47 Jens

謝謝@Jens它的工作 – Saurabh

您正在錯誤地轉義表名稱。使用這個原始查詢：

SELECT DISTINCT(t1.testName), 
     t2.comments AS C1 
FROM `sample` AS t1 
LEFT JOIN `sample` AS t2 
    ON t1.testName = t2.testName 
WHERE t1.buildNumber = 181 AND 
     t2.buildNumber = 180 AND 
     t1.errorStackTrace IS NOT NULL

我不認爲你真的需要任何反引號。但是在任何情況下，只有列名稱需要反撥，而不是別名，例如，

t1.`testName` but NOT `t1.testName`

來源

2016-07-28 05:22:14

錯反引號，試試這個：

$sql3 = "SELECT distinct(t1.`testName`), t2.`comments` AS C1 from `sample` AS t1 left join `sample` AS t2 ON t1.`testName`= t2.`testName` where t1.`buildNumber`= 181 and t2.`buildNumber`= 180 and t1.`errorStackTrace` is not null";

來源

2016-07-28 05:22:14 Blank

SQL錯誤在PHP返回的查詢與聯接

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