blocks|key|699076|text|尝试修剪字符串以确保没有额外的空格：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|699077|Cursor+c+=+db.rawQuery("SELECT+*+FROM+tbl1+WHERE+TRIM(name)+=+'"%2Bname.trim()%2B"'",+null);|code-block|syntax|javascript|699078|也可以使用c.moveToFirst()，比如前面提到的@thinksteep。|offset|length|style|CODE|699079|entityMap^0|0|0|5|F|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Try trimming the string to make sure there is no extra white space:

<pre><code>Cursor c = db.rawQuery("SELECT * FROM tbl1 WHERE TRIM(name) = '"+name.trim()+"'", null);
</code></pre>

Also use <code>c.moveToFirst()</code> like @thinksteep mentioned.

blocks|key|1602062|text|这是select语句的完整代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1602063|SQLiteDatabase+db+=+this.getReadableDatabase();
Cursor+c+=+db.rawQuery("SELECT+column1,column2,column3+FROM+table+",+null);
if+(c.moveToFirst()){
++++do+{
++++++++//+Passing+values+
++++++++String+column1+=+c.getString(0);
++++++++String+column2+=+c.getString(1);
++++++++String+column3+=+c.getString(2);+
++++++++//+Do+something+Here+with+values
++++}+while(c.moveToNext());
}
c.close();
db.close();|code-block|syntax|javascript|1602064|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is a complete code for select statements.

<pre><code>SQLiteDatabase db = this.getReadableDatabase();
Cursor c = db.rawQuery("SELECT column1,column2,column3 FROM table ", null);
if (c.moveToFirst()){
 do {
 // Passing values 
 String column1 = c.getString(0);
 String column2 = c.getString(1);
 String column3 = c.getString(2); 
 // Do something Here with values
 } while(c.moveToNext());
}
c.close();
db.close();
</code></pre>

blocks|key|4736627|text|尝试使用以下语句：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4736628|Cursor+c+=+db.rawQuery("SELECT+*+FROM+tbl1+WHERE+name+=+?",+new+String[]+{name});|offset|length|style|CODE|4736629|Android要求WHERE子句与a+?进行比较，然后指定相同数量的？在查询的第二个参数中(当前有null)。|4736630|正如Nambari提到的，您应该使用c.moveToFirst()而不是c.moveToNext()|4736631|还有，名字里有没有引用？这也可能会把事情搞砸。|4736632|entityMap^0|0|0|29|0|0|I|F|10|E|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@$D|R|E|S|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|T|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|U|8|@$D|V|E|W|F|G]|$D|X|E|Y|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|Z|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|O|$]]

Try using the following statement:

<code>Cursor c = db.rawQuery("SELECT * FROM tbl1 WHERE name = ?", new String[] {name});</code>

Android requires that WHERE clauses compare to a ?, and you then specify an equal number of ? in the second parameter of the query (where you currently have null).

And as Nambari mentioned, you should use <code>c.moveToFirst()</code> rather than <code>c.moveToNext()</code>

Also, could there be quotations in name? That could screw things up as well.

blocks|key|4736640|text|下面是下面的代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4736641|String+areaTyp+=+"SELECT+"+%2BAREA_TYPE+%2B+"++FROM+"
++++++++++++%2B+AREA_TYPE_TABLE+%2B+"+where+`"+%2B+TYPE+%2B+"`="
++++++++++++%2B+id;|code-block|syntax|javascript|4736642|其中id是显示结果的条件。|4736643|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Here is the below code.

<pre><code>String areaTyp = "SELECT " +AREA_TYPE + " FROM "
 + AREA_TYPE_TABLE + " where `" + TYPE + "`="
 + id;
</code></pre>

where id is the condition on which result will be displayed.

blocks|key|699177|text|+public+User+searchUser(String+name)+{
++++User+u+=+new+User();
++++SQLiteDatabase+db+=+this.getWritableDatabase();+//get+the+database+that+was+created+in+this+instance
++++Cursor+c+=+db.rawQuery("select+*+from+"+%2B+TABLE_NAME_User%2B"+where+username+=?",+new+String[]{name});
++++if+(c.moveToLast())+{
++++++++u.setUsername(c.getString(1));
++++++++u.setEmail(c.getString(1));
++++++++u.setImgUrl(c.getString(2));
++++++++u.setScoreEng(c.getString(3));
++++++++u.setScoreFr(c.getString(4));
++++++++u.setScoreSpan(c.getString(5));
++++++++u.setScoreGer(c.getString(6));
++++++++u.setLevelFr(c.getString(7));
++++++++u.setLevelEng(c.getString(8));
++++++++u.setLevelSpan(c.getString(9));
++++++++u.setLevelGer(c.getString(10));
++++++++return+u;

++++}else+{
++++++++Log.e("error+not+found",+"user+can't+be+found+or+database+empty");
++++++++return+u;
++++}

}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|699178|这是我的代码，用来选择一个用户并且只有一个，所以你初始化一个空的对象，然后调用你的可写数据库，使用一个游标，以防有很多人，你需要一个游标，在这里，你可以选择使用:+1-|unstyled|offset|length|style|BOLD|699179|+++if+(c.moveToLast())+{+}+//it+return+the+last+element+in+that+cursor+|699180|或使用:+2-|699181|+if(c.moveToFirst())+{++}+//return+the+first+object+in+the+cursor+|699182|别忘了，如果数据库是空的，你必须处理这个问题，在我的例子中，我只是返回一个空对象|699183|好运|699184|entityMap^0|0|0|L|0|0|0|0|0|0|2|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|X|8|@$G|Y|H|Z|I|J]]|9|@]|A|$]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$B|C]]|$1|M|3|N|5|F|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$B|C]]|$1|Q|3|R|5|F|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@$G|15|H|16|I|J]]|9|@]|A|$]]|$1|U|3|-4|5|F|7|17|8|@]|9|@]|A|$]]]|V|$]]

<pre><code> public User searchUser(String name) {
 User u = new User();
 SQLiteDatabase db = this.getWritableDatabase(); //get the database that was created in this instance
 Cursor c = db.rawQuery("select * from " + TABLE_NAME_User+" where username =?", new String[]{name});
 if (c.moveToLast()) {
 u.setUsername(c.getString(1));
 u.setEmail(c.getString(1));
 u.setImgUrl(c.getString(2));
 u.setScoreEng(c.getString(3));
 u.setScoreFr(c.getString(4));
 u.setScoreSpan(c.getString(5));
 u.setScoreGer(c.getString(6));
 u.setLevelFr(c.getString(7));
 u.setLevelEng(c.getString(8));
 u.setLevelSpan(c.getString(9));
 u.setLevelGer(c.getString(10));
 return u;

 }else {
 Log.e("error not found", "user can't be found or database empty");
 return u;
 }

}
</code></pre>

this is my code to select one user and one only
so you initiate an empty object of your class then
 you call your writable Database 
use a cursor in case there many and you need one 
here you have a choice Use : 1-

<pre><code> if (c.moveToLast()) { } //it return the last element in that cursor 
</code></pre>

or Use : 2- 

<pre><code> if(c.moveToFirst()) { } //return the first object in the cursor 
</code></pre>

and don't forget in case the database is empty you'll have to deal with that in my case i just return an empty object 

Good Luck

blocks|key|699223|text|详细答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|699224|String[]+selectionArgs+=+new+String[]{name};
++++Cursor+c+=+db.rawQuery(
+++++++"SELECT+*+FROM+"+%2B+tabl1+%2B+
+++++++"+WHERE+"+%2B+name+%2B+"+=+?+",+selectionArgs
++++);|code-block|syntax|javascript|699225|selectionArgs+:将您想要与之进行比较的'name'作为参数。|offset|length|style|BOLD|CODE|699226|这里请注意“一个游标对象，它位于您引用的表的第一个条目之前”。|699227|因此，要移动到第一个条目：|699228|c.moveToFirst();|699229|getColumnIndex(String+ColumnName)+:这将返回给定列名的从零开始的列索引。|699230|tv.setText(c.getString(c.getColumnIndex("email")));|699231|在这种情况下，您想要在“name”列中搜索多行，然后使用循环，如下所示：|699232|if+(cursor.moveToFirst()){
++++do{
++++++++////go+traversing+through+loops
+++++++++++
++++}while(cursor.moveToNext());
}|699233|这应该可以解决问题。|699234|entityMap^0|0|0|0|F|Q|6|0|5|Q|0|0|0|0|Z|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|17|8|@$I|18|J|19|K|L]|$I|1A|J|1B|K|M]]|9|@]|A|$]]|$1|N|3|O|5|6|7|1C|8|@$I|1D|J|1E|K|L]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1F|8|@]|9|@]|A|$]]|$1|R|3|S|5|D|7|1G|8|@]|9|@]|A|$E|F]]|$1|T|3|U|5|6|7|1H|8|@$I|1I|J|1J|K|L]]|9|@]|A|$]]|$1|V|3|W|5|D|7|1K|8|@]|9|@]|A|$E|F]]|$1|X|3|Y|5|6|7|1L|8|@]|9|@]|A|$]]|$1|Z|3|10|5|D|7|1M|8|@]|9|@]|A|$E|F]]|$1|11|3|12|5|6|7|1N|8|@]|9|@]|A|$]]|$1|13|3|-4|5|6|7|1O|8|@]|9|@]|A|$]]]|14|$]]

Detailed answer:
<pre><code>String[] selectionArgs = new String[]{name};
 Cursor c = db.rawQuery(
 &quot;SELECT * FROM &quot; + tabl1 + 
 &quot; WHERE &quot; + name + &quot; = ? &quot;, selectionArgs
 );
</code></pre>
selectionArgs : this takes the <code>'name'</code> you desire to compare with, as argrument.
Here note &quot;A Cursor object, which is positioned before the first entry of the table you refer to&quot;.
So,to move to first entry :
<pre><code>c.moveToFirst();
</code></pre>
getColumnIndex(String ColumnName) : this returns the zero-based column index for the given column name.
<pre><code>tv.setText(c.getString(c.getColumnIndex(&quot;email&quot;)));
</code></pre>
In case, you want to go searching through multiple rows for a given name under 'name' column then use loop as below:
<pre><code>if (cursor.moveToFirst()){
 do{
 ////go traversing through loops
 
 }while(cursor.moveToNext());
}
 
</code></pre>
This should solve the problem.

I have taken <code>String</code> value from a <code>EditText</code> and set it inside <code>SELECT QUERY after WHERE</code> condition

As

<pre><code>TextView tv = (TextView) findViewById(R.id.textView3);
EditTextet2 et = (EditText) findViewById(R.id.editText1);

String name = et.getText().toString();

Cursor c = db.rawQuery("SELECT * FROM tbl1 WHERE name = '"+name+"'", null); 

c.moveToNext();

tv.setText(c.getString(c.getColumnIndex("email")));
</code></pre>

But it doesn't work. Any suggestions?

Android SQLite SELECT Query

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我从EditText中获取了String值，并将其设置在SELECT QUERY after WHERE条件中作为TextView tv = (TextView) findViewById(R.id.textView3);EditTextet2 et = (EditText) findViewById(R.id.editText1);String name = et.getText().toStr

问Android SQLite SELECT查询
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回答 6

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问Android SQLite SELECT查询
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