Php更新qeury无法正常工作 [英] Php update qeury not working without error

问题描述

每个人都有项目从哪里获取价值，然后在另一张桌子上更新，这里的代码我觉得代码很好，但我不知道为什么它不起作用：

选择代码正在运行它只是更新



我尝试过：

hi every one i have project where to get value from table and then update it on the another table, here is the code i think the code is good but i don't know why it's not working :
the Selection code is working it's just the updating

What I have tried:

if($server)
	{
				echo "serve done<br>";
		
				$MutashabehatTable2='SELECT * FROM mutashabehat WHERE SurahNumber=2 AND AyahNumber=1 AND SurahNumber2=13 AND AyahNumber2=1';
				
				$sql = mysqli_query($server,$MutashabehatTable2 );// get Fadwa and sumaih mutashabeha table
				
				$numpOfRows2 = mysqli_num_rows($sql); // similar verses number of rows
			echo $numpOfRows2."<br>";
			//If there are similar verses 
			if( $numpOfRows2 > 0)
		{
			
				$row = mysqli_fetch_assoc($sql);
							$DifferentParts1= "". $row['DifferentParts1'];
				$DifferentParts2= "". $row['DifferentParts2'];
				$mark= "".$row['Markers'];
				echo $MutashabehatTable1='UPDATE \'mutashabeha\' SET SimilarParts = \''.$row['SimilarParts'].'\', DifferentParts1 = \'$DifferentParts1\', DifferentParts2 =  \'$DifferentParts2\' , Markers = \'$mark\' WHERE SurahNumber=\'2\' AND AyahNumber= \'1\' AND SurahNumber2= \'13\' AND AyahNumber2= \'1\' ';
				
				$sql2 = mysqli_query($server,$MutashabehatTable1 );// get Fadwa and sumaih mutashabeha table
				if($sql2)
					echo 'done';
				else
	echo mysqli_error();
			die;			
		}//end row number if
	}//end if connection done
	//If conn Failed 
	else{
				die("connection failed: ".mysqli_connect_error());

	}

推荐答案

服务器）
{
echo 服务完成< br>;

MutashabehatTable2 = ' SELECT * FROM mutashabehat WHERE SurahNumber = 2 AND AyahNumber = 1 AND SurahNumber2 = 13 AND AyahNumber2 = 1' ;

MutashabehatTable2='SELECT * FROM mutashabehat WHERE SurahNumber=2 AND AyahNumber=1 AND SurahNumber2=13 AND AyahNumber2=1';

sql = mysqli_query（

sql = mysqli_query(

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