SQL Error returned in php for query with joins

Question

I am trying to populate a table by executing the below query for a php application:

$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));
}

I am seeing the following error returned:

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

I have tried a lot to fix this but couldn't. The query runs fine when run on mysql. Any help would be appreciated. Thanks

use proper back quotes or remove $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"; — Rakesh Sharma
– Rakesh Sharma, Commented Jul 28, 2016 at 5:21

Jens · Accepted Answer · 2016-07-28 05:22:47Z

1

You have to use the backticks only arround table names or column names not including the alias:

$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";

answered Jul 28, 2016 at 5:22

Jens

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Comments

Tim Biegeleisen · Accepted Answer · 2016-07-28 05:22:14Z

1

You are escaping the table name incorrectly. Use this raw query:

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

I don't think you really need backticks anywhere. But in any case, only column names need to be backticked, never aliases, e.g.

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

answered Jul 28, 2016 at 5:22

Tim Biegeleisen

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Comments

Blank · Accepted Answer · 2016-07-28 05:22:14Z

0

Wrong back quote, try this:

$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";

answered Jul 28, 2016 at 5:22

Blank

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